12 g CH3OH are produced when 14 grams of carbon monoxide reacts with 1.5 grams of hydrogen gas.
Writing the balanced equation for the reaction taking place:
CO(g) + 2H2(g) ⇒ CH3OH
Finding the limiting reactant:
moles CO present ⇒ 14 g x 1 mol CO / 28 g = 0.5 moles CO
moles H2 present ⇒ 1.5 g x 2 mol H2 / 2 g = 0.75 moles H2O
Because according to the balanced equation it takes 2 moles of H2 for every 1 mole CO, there isn't enough H2, so it is limiting and will dictate how much CH3OH can be made.
Mass of CH3OH produced
= 0.75 moles H2 x 1 mol CH3OH / 2 mol H2 x 32 g CH3OH / mol
= 12 g CH3OH
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