WBBSE Class 10 Physical Science Chapter 3 Chemical Calculations Long Answer Questions

Class 10 Physical Science WBBSE Chapter 3 Chemical Calculations Broad Answer Type Questions

Question 1. What steps should be followed to solve a problem based on the chemical equation by the weight-weight method?
Answer:

The following steps are followed :

• First, write down the complete balanced chemical equation.
• The molecular weight of the reactants and products is to be calculated from their respective formulae by adding the atomic weights of the concerned elements.
• The unknown weight of the substance asked for in the question is to be calculated from the calculated weight in the equations
• The same units are to be used for all quantities.

Question 2. What is the information available from the formula CaCO₃?
Answer:

The formula CaCO₃ conveys the following information :

• It stands for calcium carbonate.
• It tells that calcium carbonate is composed of calcium, carbon, and oxygen elements.
• It represents a molecule of calcium carbonate.
• It represents that a molecule of calcium carbonate is made up of one atom each of calcium and carbon and three atoms of oxygen.
• It stands for 100 parts by weight of calcium carbonate i.e. its molecular weight and signifies that the ratio of calcium, carbon, and oxygen by weight in it is 40:12:48.

Question 3. What is meant by the balancing of a chemical equation? Why it is necessary?
Answer:

Balancing of a chemical equation: It means making both sides of the equation equal with respect to the kind and number of the atoms of the elements involved.

Balancing of a chemical equation is necessary: Due to the conservation of mass and indestructibility of matter, no atoms can be created or destroyed in a chemical reaction Hence, The same kind of atoms in the same numbers must be present on both sides of the chemical equation.

Question 4. Write the following as balanced equations.

Cu+HNO₃ → Cu (N0₃) + O₂

Pb (N0₃)₂ → PbO + NO₂ + O₂

Al + NaOH + H₂O → NaAlO₂ + H₂

CuO + NH₃ → Cu + N₂ + H₂O

Answer:

Balanced equations are:

3Cu + 8HNO₃ = 3Cu (NO₃)₂  + 2NO – 4H₂O

2Pb (NO₃)₂  = 2pb0+ 4NO₂ +O₂

2Al + 2NaOH + 2H₂O = 2NaAIO₂ + 3H₂

2NH₃ +3CuO = 3Cu + 2n₂ + 3H₂0

Question 5. What are the limitations of a chemical equation?
Answer:

Limitation of a chemical equation :

Under what conditions a chemical reaction occurs i.e. pressure, temperature, catalyst are not-known.

• A reaction whether exothermic or endothermic is not known from a chemical equation.
• The chemical equations cannot provide information about the completion of the reaction or the attainment of equilibrium.
• A chemical equation does not give any idea about the reversibility of the reaction.
• From the chemical equation, the time required for the completion of the equation is not known.
• A reaction whether slow or fast is not known from the equation.
• The equation does not provide information regarding the nature of the reactants and products viz. solid, liquid, or gas.

Question 6. Balance the equation KCI0₃ → KCI + O₂ by trial and error method.
Answer:

KCI0₃ KCI + O₂: It is seen that the number of atoms of oxygen on the left-hand side is 3 whereas in the right-hand side it is 2. To equalize the number of atoms of oxygen on both sides it requires to multiply KCIO₃ by 2 and oxygen by 3.

2KCIO₃ → KCI +30₂ SO as to balance this equation if KCI is multiplied by then the equation is proper balance the equation if KCI is multiplied by 2 then the equation is proper. balanced.

2KCIO₃ = 2KCI + 30₂

Question 7. What information is obtained from the chemical equations?
Answer:

A chemical equation gives the following information:

1. Qualitative informations:

• From the chemical equation, the naming of elements and compounds taking part in the reaction and also the products are known.
• From the chemical equation, the symbols and the formulae of the reactants and products are also known.

2. Quantitative information:

• A number of atoms or molecules of the reactants and products involved in the reaction are known.
• How many parts of elements and compounds take part in the reaction and production is also known.
• If the reactants and products are all gaseous then at the same temperature and pressure. The ratio in volumes is known.

Question 8. Explain the method of balancing a chemical equation by trial and error method.
Answer:

Balancing of a chemical equation by trial and error method :

• In the method, the proper numbers are put before reactants and products so that the number of atoms on both sides is equal.
• In doing so, we must be cautious to see that all the reactants and products exist as molecules and not as atoms as the free existence of atomic is not possible.

Class 10 Physical Science WBBSE

Question 9. What is the method of writing chemical equations?
Answer:

Method of writing chemical equation :

Initially, the symbols of atoms of elements and formulae of molecules of reactants and products are written.

• Symbols and formulae of the reactants are written on the left side and those for products are on the right side. An arrow (→) is placed between reactants and products. For more than one reactants and product a plus sign (+) is to be given between the reactants and also between the products.
• The number of atoms of the reactants on the left-hand side must be equal to the number of atoms of the products on the right-hand side so as to maintain the law of conservation of mass. To equalize both sides, proper multiplication is required so that the number of atoms on both sides are same.
• Now the equation is expressed by replacing the arrow sign with a sign of equal (=).

Question 10. What information is obtained from the equation: 3H₂ + N₂ = 2NH₂
Answer:

The following information is obtained from the equation: 3H₂+N₂ = 2NH₂

Qualitative information: The reactants are hydrogen and nitrogen, and ammonia is the product.

Quantitative information :

• Three molecules of hydrogen combine with one molecule of nitrogen to produce two molecules of ammonia.
• Three volumes of hydrogen and one volume of nitrogen combine to produce two volumes of ammonia at the same temperature and pressure and their ratio is 3:1:2.
• Three moles of hydro and one mole of nitrogen combine chemically to produce two moles of ammonia.

Question 11. What information is obtained from the equation? C + O₂ = CO₂
Answer:

The following information is obtained from an equation: C + O₂= CO₂

Qualitative informations :

• From the above equation, it is known that carbon combines with oxygen producing carbon dioxide.
• Their respective symbols and formulas are also known

Quantitative information:

• One carbon atom combines with one oxygen molecule producing one carbon dioxide molecule.
• Hence the total number of atoms on the left-hand side is (1+2) = 3 and the number of atoms on the right-hand side is also 3.
• One gram-atom carbon reacts with one gram-molecule of oxygen giving rise to one gram-molecule of carbon dioxide.
• 12g carbon combines with 32g oxygen yielding 44g carbon dioxide.
• Hence, the total mass on the left-hand side (12+32) g = 44g is equal to the total mass on the right-hand side 44g. It proves the law of conservation of mass.
• Again 12g carbon combines with 32g oxygen producing 22.4 litre of carbon dioxide at STP.

Question 12. How many grams of oxygen evolve when 122.5g potassium chlorate is heated? (Given K = 39, CI = 35.5 0 = 16)
Answer:

The balanced equation is : \(\begin{array}{ll}
2 \mathrm{KClO}= & 2 \mathrm{KCl}+3 \mathrm{O}_2 \\
2(39+35.5+3 \times 161 \mathrm{~g} & 3[2 \times 16] \mathrm{g} \\
=245 \mathrm{~g} & =96 \mathrm{~g}
\end{array}\)

By heating 245g KCIO₃ 96g O₂ is obtained

∴ By heating 122.5g KCIO₃ \(\frac{96 \times 122.5}{245} \mathrm{~g}\) is obtained.

Physics Class 10 WBBSE

Question 13. 2.6g zinc is treated with excess dil H₂SO₄ How many grams of oxygen combine with the evolved hydrogen?
Answer:

The balanced equation for the production of hydrogen is:

\(\underset{65 \mathrm{~g}}{\mathrm{Zn}}+\mathrm{H}_2 \mathrm{SO}_4=\mathrm{ZnSO}_4+\underset{(2 \times 1) \mathrm{g}=2 \mathrm{~g}}{\mathrm{H}_2}\)

So, 65g of Zinc produces 2g of hydrogen

∴ 2-6 Zinc produces \(\frac{2 \times 2.6}{65} \mathrm{~g}=0.089\) hydrogen

Now, the reaction where hydrogen and oxygen combine is: 2H₂ + O₂ = 2H₂O

So, 4g hydrogen combined with 32g oxygen

0.08g hydrogen combined with \(\frac{32 \times 0.08}{4} \mathrm{~g}=0.64 \mathrm{~g}\)

Question 14. What is the observed loss in weight of 5g? calcium carbonate when it undergoes thermal decomposition?
Answer:

The balanced equation is : \(\begin{aligned}
& \mathrm{CaCO}_3=\mathrm{CaO}+\mathrm{CO}_2 \\
& (40+12+16 \times 3) \mathrm{g}(12+16 \times 2) \mathrm{g} \\
& =100 \mathrm{~g} \quad=44 \mathrm{~g} \\
&
\end{aligned}\)

Loss in weight in the weight of CO₂ that escapes 100g CaCO₃ produces 44g CO₂

∴ 5g CaCO₃ produces \(\frac{44 \times 5}{100} \mathrm{~g}=2.2 \mathrm{gCO}_2\)

Question 15. What is the percentage of ammonia in that quantity of ammonium chloride that can produce 5g? ammonia?
Answer:

The balanced equation is :

\(\begin{aligned}
& 2 \mathrm{NH}_4 \mathrm{Cl}+\mathrm{CaO}=2 \mathrm{NH}_3+\mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O} \\
& 2[14+1 \times 4+35.5)_{\mathrm{g}} 2[14+1 \times 3]_{\mathrm{g}} \\
& =107 \mathrm{~g}
\end{aligned}\)

So, 34g ammonia is obtained from 107g NH₂CI.

Question 16. What weight of potassium chlorate of 96% purity will yield 4.8g oxygen on complete thermal decomposition? (Given, K = 39, CI = 35.5, O = 16)
Answer:

The balance equation is: 2KCIO₃ (245g) = 2KCI + 3O₂(96g)

∴ 96g oxygen is obtained from 245g KCIO₂ of 100% purity.

∴ 4.8g oxygen is obtained from \(\frac{245 \times 4.8}{96} \mathrm{~g}\)

Let x gram of 96% purity contain 12.25g KCIO₃ of 100% parity

∴ \(\frac{96}{100} \times 12.56 \text { or, } x=\frac{12.25 \times 100}{96}=12.76 \text { (approx) }\)

So, the required quantity of KCIO₃ = 12.76g

Physics Class 10 WBBSE

Question 17. On strong heating, limestone decomposes into quicklime and carbon dioxide. How much quantity of limestone will produced on complete decomposition, 30g of quicklime by the above reaction?
Answer:

The balanced equation is: \(\begin{array}{ll}
\mathrm{CaCO}_3=\mathrm{CaO}+\mathrm{CO}_2 & \\
(40+12+3 \times 16) \mathrm{g} & {[40+16] \mathrm{g}} \\
=100 \mathrm{~g} & =56
\end{array}\)

So 56g, CaO is obtained by the complete decomposition of 100g CaCO₃

∴ 30g CaO is obtained by the complete decomposition of \(\frac{100 \times 30}{56} \mathrm{~g} \mathrm{CaCO}_3=53.6 \mathrm{~g} \mathrm{CaCO}_3\)

Thus 53.6g of limestone will have to be decomposed.

Question 18. How many grams of magnesium metal will give 1.2g hydrogen in a complete reaction with dilute H₂SO₄ (Mg= 24, H = 1) The balanced equation is:
Answer:

\(\begin{aligned}
\mathrm{Mg}+\mathrm{H}_2 \mathrm{SO}_4= & \mathrm{Mg} \mathrm{SO}_4+\mathrm{H}_2 \\
24 \mathrm{~g} & (1 \times 2) \mathrm{g}=2 \mathrm{~g}
\end{aligned}\)

So 2g H₂ is obtained from 24g Mg

∴ 1.2g H₂ is obtained from \(\frac{24 \times 1.2}{2} \mathrm{~g} \mathrm{Mg}=14.4 \mathrm{gMg} .\)

So, 14.4g will be required

Question 19. Calculate the number of moles in 1.5 gm of ammonia.
Answer:

We know, number of mole = \(\frac{\text { Mass given }}{\text { gram molecular mass }}\)

\(\begin{aligned}
& =\frac{1.5}{1.7} \\
& =0.88
\end{aligned}\)

Question 20. The vapor density of a gas is 40. Calculate the volume of 20 gm of this gas at 27°C and 950 mm pressure.
Answer:

Given

The vapor density of a gas is 40.

The molecular weight of the gas = 2 x vapor density

= 2 × 40 = 80

The volume of 80 gm of gas at S.T.P. is 22.4 Lit

The volume of 10 gm of gas at S.T.P.

\(=\frac{22.4 \times 10}{80}=2.8 \mathrm{Lit}\)

Let the volume of this 2.8 liter of gas at 27°C and 950 mm pressure be V Litre

\(\frac{\mathrm{V} \times 950}{273+27}=\frac{2.8 \times 760}{273}\)  ⇒V = 7.79 Litre

Hence the volume of the gas at 27°C and 950 pressure is 4.9 Litre.