QuestionApril 23, 2025

How much methanol (CH_(3)OH, in grams) can be formed from 43.2 g of hydrogen? Assume excess CO. CO(g)+2H_(2)(g)arrow CH_(3)OH(g) square gCH_(3)OH

How much methanol (CH_(3)OH, in grams) can be formed from 43.2 g of hydrogen? Assume excess CO. CO(g)+2H_(2)(g)arrow CH_(3)OH(g) square gCH_(3)OH
How much methanol (CH_(3)OH, in grams) can be formed from 43.2 g of hydrogen? Assume excess CO.
CO(g)+2H_(2)(g)arrow CH_(3)OH(g)
square gCH_(3)OH

Solution
4.0(144 votes)

Answer

342.64 g CH_3OH Explanation 1. Calculate moles of hydrogen Use the molar mass of H_2: **M(H_2) = 2.02 \, \text{g/mol}**. Moles of H_2 = \frac{43.2 \, \text{g}}{2.02 \, \text{g/mol}} = 21.39 \, \text{mol}. 2. Determine moles of methanol produced From the balanced equation, 2 moles of H_2 produce 1 mole of CH_3OH. Thus, moles of CH_3OH = \frac{21.39}{2} = 10.695 \, \text{mol}. 3. Calculate grams of methanol Use the molar mass of CH_3OH: **M(CH_3OH) = 32.04 \, \text{g/mol}**. Grams of CH_3OH = 10.695 \, \text{mol} \times 32.04 \, \text{g/mol} = 342.64 \, \text{g}.

Explanation

1. Calculate moles of hydrogen<br /> Use the molar mass of $H_2$: **$M(H_2) = 2.02 \, \text{g/mol}$**. Moles of $H_2 = \frac{43.2 \, \text{g}}{2.02 \, \text{g/mol}} = 21.39 \, \text{mol}$.<br /><br />2. Determine moles of methanol produced<br /> From the balanced equation, 2 moles of $H_2$ produce 1 mole of $CH_3OH$. Thus, moles of $CH_3OH = \frac{21.39}{2} = 10.695 \, \text{mol}$.<br /><br />3. Calculate grams of methanol<br /> Use the molar mass of $CH_3OH$: **$M(CH_3OH) = 32.04 \, \text{g/mol}$**. Grams of $CH_3OH = 10.695 \, \text{mol} \times 32.04 \, \text{g/mol} = 342.64 \, \text{g}$.
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