(solved)Question 1.4 of NCERT Class XI Chemistry Chapter 1

Molar mass of $\displaystyle \text{CO}_2$ = 12 + 2 x 16 = 44 g/mol

Write the balanced equation first

$\displaystyle \text{C (s)} + \text{O}_2\text{ (g)} \rightarrow \text{CO}_2\text{ (g) . . . (1)}$

$\displaystyle \underline{\underline{\text{Part (i)}}}$

Given: 1 mol carbon + excess dioxygen

by (1) $\displaystyle 1 \text{ mol C will give } 1 \text{ mol CO}_2$

but molar mass of CO2 is 44 g/mol

$\displaystyle \therefore 1 \text{ mol CO}_2 \text{ produced } \equiv 44\text{ g CO}_2\:\underline{Ans} $

$\displaystyle \underline{\underline{\text{Part (ii)}}}$

Given: 16 g dioxygen ≡ 1/2 mol $\displaystyle \text{O}_2$

Also given: 1 mol carbon

by (1) $\displaystyle 1\text{ mol C reacts }\equiv 1 \text{ mol O}_2$

but we have only 1/2 mol dioxygen

$\displaystyle \text{hence 1/2 mol dioxygen is limiting reagant}$

Adjust the equation:

$\displaystyle \underset{1/2 \text{ mol}}{\text{C (s)}} + \underset{1/2 \text{ mol}}{\text{O}_2\text{ (g)}} \rightarrow \underset{1/2 \text{ mol}}{\text{CO}_2\text{ (g)}}$

$\displaystyle \implies \frac12 \text{ mol CO}_2 $ produced

but molar mass of CO2 is 44 g/mol

$\displaystyle \therefore 1/2 \times 44 = 22 \text{ g CO}_2 \:\underline{Ans} $

$\displaystyle \underline{\underline{\text{Part (iii)}}}$

dioxygen, being less, will still be the limiting reagant

Hence, the answer remains same $\displaystyle 22 \text{ g CO}_2 \:\underline{Ans}$

First write the balanced chemical equation.

Then convert grams to moles for all reactants and products.

Find the limiting reagant - the one that will be consumed first.

Adjust the multipliers according to the moles of the limiting reagant.