Metallic magnesium reacts with steam to produce magnesium hydroxide and hydrogen gas. If 16.2 g of #Mg# are heated with 12.0 g of #H_2O#, what is the limiting reactant?

Question

Henry Antrim · Accepted Answer

Water.

Balanced equation of the reaction is #Mg+2H_2O->Mg(OH)_2+H_2 uarr# Molar masses of the elements in the reactants: 24.3050 for magnesium, 1.00794 for hydrogen, and 15.9994 for oxygen Given 12 gm of Water #=12.0/(2xx1.00794+15.9994)approx# 0.6661 moles And 16.2 g of Magnesium #=16.2/24.3050approx# 0.6665 moles We observe from the equation that #1 # mol of magnesium requires #2# mols of water. We require #2xx0.6665=1.333# moles of water to completely consume given quantity of magnesium. We have only 0.6661 moles. Therefore, water is the limiting reactant.

Luna Chen · Answer

undefined To determine the limiting reactant, we need to compare the amount of product that each reactant can produce. We first need to calculate the moles of each reactant using their respective molar masses. Then, we use the stoichiometry of the reaction to find out how many moles of the product each reactant can produce. Whichever reactant produces the lesser amount of product is the limiting reactant.

1. Calculate the moles of Mg: $ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $
2. Calculate the moles of H2O: $ \text{moles} = \frac{\text{mass}}{\text{molar mass}} $
3. Use the balanced chemical equation to find the theoretical yield of Mg(OH)2 for each reactant.
4. The limiting reactant is the one that produces the lesser amount of product.

However, since we don't have the balanced chemical equation for the reaction, we can't proceed with the calculation. Could you provide the balanced equation for the reaction?