Question

$2240$ ml of $H_2$ at NTP will be given by reaction of:

• A. $4.6$ g of $Na$ with excess of water
• B. $6.5$ g of $Zn$ with $5$ g of $HCl$
• C. $6.35$ g of $Cu$ with excess of dil. $HCl$
• D. All of the above

Answer 1

The correct option is A $4.6$ g of $Na$ with excess of water

At NTP, $2240$ ml of hydrogen corresponds to $\frac{2240}{22400}=0.1$ moles.

$2Na+2H_{2}O\to 2NaOH+H_2$

$2$ moles of sodium give $1$ mole of hydrogen.
$\therefore$ Number of moles of $Na$ required to liberate $0.1$ moles of $H_2$ will be $\frac{0.1\times 2}{1}=0.2$moles

$0.2$ moles of sodium (atomic weight is $23$ g/mol) corresponds to $4.6$ g.