Question

?CH3OH + ?O2 => ?HCO2H + ?H2O,& & what is the maximum amount of HCO2H(46.0254 g/mol) which could be formed from 5.61 g of CH3OH(32.0419g/mol) and 17.65 g of O2(31.9988g/mol)

Answer 1

Answer on Question #44825–Chemistry–Inorganic Chemistry

Question

$\mathrm{CH_3OH} + \mathrm{O_2} \rightarrow \mathrm{HCO_2H} + \mathrm{H_2O}$. What is the maximum amount of $\mathrm{HCO_2H}$ (46.0254 g/mol) which could be formed from 5.61 g of $\mathrm{CH_3OH}$ (32.0419 g/mol) and 17.65 g of $\mathrm{O_2}$ (31.9988 g/mol)?

Solution

Balanced chemical equation is $\mathrm{CH_3OH} + \mathrm{O_2} \rightarrow \mathrm{HCO_2H} + \mathrm{H_2O}$, so methanol and oxygen react in equimolar ratio.

Let us calculate the mass of $\mathrm{O_2}$ needed to react with 5.61 g of $\mathrm{CH_3OH}$

$$\begin{array}{c c c c c} 5.61 \, \mathrm{g} & & \mathrm{m_{O_2}}, \, \mathrm{g} \\ \mathrm{CH_3OH} & + & \mathrm{O_2} & \rightarrow & \mathrm{HCO_2H} + \mathrm{H_2O} \\ 32.0419 \, \mathrm{g/mol} & & 31.9988 \, \mathrm{g/mol} \end{array}$$ $$m_{\mathrm{O_2}} = \frac{m_{\mathrm{CH_3OH}} \cdot M_{\mathrm{O_2}}}{M_{\mathrm{CH_3OH}}} = \frac{5.61 \, \mathrm{g} \cdot 31.9988 \, \mathrm{g/mol}}{32.0419 \, \mathrm{g/mol}} = 5.60 \, \mathrm{g}$$

Thus, 5.60 g of $\mathrm{O_2}$ are needed to completely oxidized 5.61 g of methanol. Since 17.65 g of $\mathrm{O_2}$ are taken for reaction, we can state that oxygen is in excess and part of it remains unreacted. So, the amount of $\mathrm{HCO_2H}$ formed should be calculated based on the mass of methanol.

$$\begin{array}{c c} 5.61 \, \mathrm{g} & \quad \mathrm{m_{HCO_2H}}, \, \mathrm{g} \\ \mathrm{CH_3OH} + \mathrm{O_2} & \rightarrow \mathrm{HCO_2H} + \mathrm{H_2O} \\ 32.0419 \, \mathrm{g/mol} & \quad 46.0254 \, \mathrm{g/mol} \end{array}$$ $$m_{\mathrm{HCO_2H}} = \frac{m_{\mathrm{CH_3OH}} \cdot M_{\mathrm{HCO_2H}}}{M_{\mathrm{CH_3OH}}} = \frac{5.61 \, \mathrm{g} \cdot 46.0254 \, \mathrm{g/mol}}{32.0419 \, \mathrm{g/mol}} = 8.06 \, \mathrm{g}$$

Answer: 8.06 g

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