Calculate the volume of hydrogen liberated at $27 C$ and 760 mm pressure by treating 1.2 g of magnesium with excess of hydrochloric acid.

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Solution

The balanced equation is:
$M g + 2 H c l \to m g c l_{2} + H_{2}$
1 mol 1 mol
24g 22.4 liter at NTP
$24$ g of mg liberate hydrogen= $22.4$ liter
$1.2$ g mg will liberate hydrogen
$\frac{22.4}{24} \times 1.2 = 1.12$ liter
the volume of hydrogen under given condition can be calculated
by applying
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$P_{1} = 760 m m$
$T_{1} = 273$ K
$V_{1}$ = $1.12$ liter
$P_{2}$ = $760$ mm
$T_{2} = (27 + 273) = 300 K$
$V_{2}$ =?
$V_{2}$ = $\frac{760 \times 1.12}{273} \times \frac{300}{760}$ = $1.2308$

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Calculate the volume of hydrogen liberated at 27C and 760 mm pressure by treating 1.2 g of magnesium with excess of hydrochloric acid.

Calculate the volume of hydrogen liberated at $27 C$ and 760 mm pressure by treating 1.2 g of magnesium with excess of hydrochloric acid.