Question

What volume of 1.0 M hydrochloric acid will be required to produce 800 cm3 hydrogen gas by reacting with excess zinc? (a single replacement reaction)

Answer 1

$1 \ cm^3 = 1 \ ml$

Moles of $H_2$ gas in $800 \ ml$ $= \frac{800}{22400}=0.03571$ $($ As volume of $1$ mole of gas is $22400\ ml)$

Moles of $H$ atoms will be twice to that of $H_2$ .

So, moles of hydrogen$=2\times 0.03571=0.07143$

Applying mole balance over $H$ to reactant$( \ HCl \ )$ and product side $( H_2)$ ,

Molarity of $HCl$ $( M)= 1 M$

So let volume required be $V$ .

$\implies MV=0.07143$

$\implies 1.0\times V= 0.07143$

$\implies V = 0.07143 \ l = 71.43 \ ml= 71.43 \ cm^3$