Question

5.0

g of marble was added to

7.5

g dilute hydrochloric acid. After the reaction was over, it was found that

0.5

g of marble was left unused. Calculate the percentage strength of hydrochloric acid. What volume of

{CO}_{2}

measured at STP will be evolved in the above reaction?

Accepted Answer

The following balanced reaction takes place when marble reacts with $HCl$ ,

$\underset{100 g}{{CaCO}_{3}} + 2 \underset{2 \times 36.5 g = 73 g}{HCl} \to {CaCl}_{2} + H_{2} O + \underset{22400 {cm}^{3} at STP}{{CO}_{2}}$

The amount of marble reacted $=$ Total amount of marble used $-$ Unused amount of marble

$= 5.0 - 0.5 = 4.5 g$

$∵ 100 g$ marble is reacted with $73 g$ of $HCl$ .

$∴ 4.5 g$ will react $= \frac{73}{100} \times 4.5 g = 3.285 g$ of $HCl$ .

And we know that $percentage strength of HCl =$ $\frac{Weight of HCl reacted}{Total weight of HCl taken} \times 100$

$= \frac{3.285}{7.5} \times 100 = 43.8 %$

Since $100$ g marble evolves $= 22400 {cm}^{3}$ carbon dioxide

Therefore, $4.5 g$ marble will evolve $= \frac{22400}{100} \times 4.5 {cm}^{3} = 1008 {cm}^{3}$

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Some Basic Concepts of Chemistry, Exercise 17: PROBLEMS FOR PRACTICE

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Some Basic Concepts of Chemistry, Exercise 17: PROBLEMS FOR PRACTICE

Questions from S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Some Basic Concepts of Chemistry, Exercise 17: PROBLEMS FOR PRACTICE with Hints & Solutions

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