Calculate the volume of hydrogen liberated at $27^{o} C$ and 760 mm pressure, by treating 0.6g of magnesium with excess of hydrochloric acid.

2.216 L

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1.651 L

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1.850 L

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0.615 L

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Solution

The correct option is D 0.615 L
$M g + 2 H C l \to M g C l_{2} + H_{2}$

Moles of $M g = \frac{0.6}{24} = 0.025$

Moles of $H_{2} = 0.025$

Volume of $H_{2} = \frac{0.025 \times 0.0821 \times 300}{1} = 0.615 l i t r e s$

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Calculate the volume of hydrogen liberated at 27oC and 760 mm pressure, by treating 0.6g of magnesium with excess of hydrochloric acid.

The correct option is D 0.615 LMg+2HCl→MgCl2+H2Moles of Mg=0.624=0.025Moles of H2=0.025Volume of H2=0.025×0.0821×3001=0.615 litres

Calculate the volume of hydrogen liberated at $27^{o} C$ and 760 mm pressure, by treating 0.6g of magnesium with excess of hydrochloric acid.

The correct option is D 0.615 L
$M g + 2 H C l \to M g C l_{2} + H_{2}$

Moles of $M g = \frac{0.6}{24} = 0.025$

Moles of $H_{2} = 0.025$

Volume of $H_{2} = \frac{0.025 \times 0.0821 \times 300}{1} = 0.615 l i t r e s$