# Question 9dbb6

Feb 5, 2015

Magnesium can react with hot water to produce magnesium hydroxide and hydrogen gas, and it can react with hot steam to produce magnesium oxide and hydrogen gas.

Since you've made no mention on which reaction takes place, I'll show you the one with hot water. The steps are identical for the other one (although you won't have a limiting reagent in that one).

So, the balanced chemical equation is

$M g \left(s\right) + 2 {H}_{2} O \left(l\right) \to M g {\left(O H\right)}_{2} \left(s\right) + {H}_{2} \left(g\right)$

The most important tool you have at your disposal when dealing with limiting reagent problems, or with any stoichiometry problem for that matter, is the mole ratio.

Let's start with the reactants. Notice that you have a $\text{1:2}$ mole ratio between magnesium and water. What that means is that, regardless of how many moles of magnesium you have, you need twice as many moles of water.

If this is not the case, you're dealing with a limiting reagent. So, you have

$\text{16.2 g Mg" * ("1 mole")/("24.3 g") = "0.667 moles Mg}$

Now check how many moles of water you have

$\text{12.0 g H"_2"O" * ("1 mole")/("18.0 g") = "0.667 moles H"_2"O}$

Since you would have needed $0.667 \cdot 2 = \text{1.33 moles}$ of water, water will be your limiting reagent. Since the maximum number of moles of water is $\text{0.667}$, the number of moles of magnesium must be

$\text{0.667 moles water" * ("1 mole magnesium")/("2 moles water") = "0.3335 moles}$

The rest of the magnesium will be in excess because you have insufficient water

n_("excess") = "0.667" - "0.3335" = "0.3335 moles Mg"#

Now let's turn to the products. Notice that you have a $\text{1:1}$ mole ratio between both magnesium and magnesium hydroxide, and magnesium and hydrogen. This means that the number of moles produced for each of the products will be equal to the moles of magnesium that react. So,

${n}_{M g {\left(O H\right)}_{2}} = \text{0.3335 moles}$ and

${n}_{{H}_{2}} = \text{0.3335 moles}$

This means that this reaction will produce

$\text{0.3335 moles" * ("58.3 g")/("1 mole") = "19.4 g}$ $M g {\left(O H\right)}_{2}$ and

$\text{0.3335 moles" * ("2.0 g")/("1 mole") = "0.667 g}$ ${H}_{2}$.