Answer:
B. 50%
Explanation:
First we convert the given masses of the reactants into moles, using their respective molar masses:
- 4 g H₂ ÷ 2 g/mol = 2 mol H₂
- 25 g CO ÷ 28 g/mol = 0.893 mol CO
0.893 moles of CO would react completely with (0.893 * 2) 1.786 moles of H₂. As there are more H₂ moles than that, H₂ is the reactant in excess and CO is the limiting reactant.
Now we calculate how many CH₃OH moles would have been formed if all CO would have been consumed:
- 0.893 mol CO * [tex]\frac{1molCH_3OH}{1molCO}[/tex] = 0.893 mol CH₃OH
Then we convert 0.893 moles of CH₃OH into grams, using its molar mass:
- 0.893 mol CH₃OH * 32 g/mol = 28.57 g
Finally we calculate the percent yield:
- 14 g / 28.57 g * 100% = 49%
