What mass of methanol (CH3OH) is produced when 86.04 g of carbon monoxide reacts with 14.14 g of hydrogen?CO(g) + 2H2(g) CH3OH(l)A.7.07 gB.112.14 gC.96.12 gD.84.06 g

To solve this problem, we need to use stoichiometry. Here are the steps:

1. First, we need to calculate the molar mass of each reactant and product. The molar mass of CO is approximately 28.01 g/mol, the molar mass of H2 is approximately 2.02 g/mol, and the molar mass of CH3OH is approximately 32.04 g/mol.

2. Next, we need to convert the given mass of each reactant to moles. For CO, we have 86.04 g / 28.01 g/mol = 3.07 moles. For H2, we have 14.14 g / 2.02 g/mol = 7.00 moles.

3. According to the balanced chemical equation, one mole of CO reacts with two moles of H2 to produce one mole of CH3OH. Therefore, the limiting reactant is CO because we have less moles of it (3.07 moles) than required to react with all the H2 (7.00 moles).

4. Since the reaction produces one mole of CH3OH for every one mole of CO, we can say that the reaction will produce 3.07 moles of CH3OH.

5. Finally, we convert the moles of CH3OH to grams using its molar mass. We have 3.07 moles * 32.04 g/mol = 98.36 g.

So, the mass of methanol produced is approximately 98.36 g. However, this option is not available in the choices given. There might be a mistake in the problem or the answer choices.