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Standard XI

Chemistry

Mole Concept

When 0.05 mol...

Question

When 0.05 moles of $H_{2}$ is mixed with $C O$ , what is the number of moles of methanol $(C H_{3} O H)$ that can be obtained? ( $C H_{3} O H$ is the only product)

0.1 mol

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0.05 mol

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0.025 mol

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0.0125 mol

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Solution

The correct option is C $0.025 mol$
$C O (g) + 2 H_{2} (g) \to C H_{3} O H (l)$
2 mol of $H_{2}$ produce 1 mol of $C H_{3} O H$
Therefore, 0.05 mol of $H_{2}$ will produce $(\frac{1 \times 0.05}{2}) = 0.025 mol of C H_{3} O H$

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When 0.05 moles of H2 is mixed with CO, what is the number of moles of methanol (CH3OH) that can be obtained? (CH3OH is the only product)

The correct option is C 0.025 molCO(g)+2H2(g)→CH3OH(l) 2 mol of H2 produce 1 mol of CH3OH Therefore, 0.05 mol of H2 will produce (1×0.052)=0.025 mol of CH3OH

When 0.05 moles of $H_{2}$ is mixed with $C O$ , what is the number of moles of methanol $(C H_{3} O H)$ that can be obtained? ( $C H_{3} O H$ is the only product)

The correct option is C $0.025 mol$
$C O (g) + 2 H_{2} (g) \to C H_{3} O H (l)$
2 mol of $H_{2}$ produce 1 mol of $C H_{3} O H$
Therefore, 0.05 mol of $H_{2}$ will produce $(\frac{1 \times 0.05}{2}) = 0.025 mol of C H_{3} O H$