Stoichiometry question? 10 points for the best answer!?

Question

Metallic magnesium reacts w/ water to produce magnesium hydroxide and hydrogen gas; if you have 16.2 g of magnesium and 12.0 g of H2O, how many moles of the excess reactant are left?

Answer 1

First get the balanced equation:

Mg + 2 H2O ---> Mg(OH)2 + H2

Now find the number of moles of each reactant; divide the mass by molar mass:

16.2g/24.3g/mole = 0.667 mole Mg

12.0g/(16.0 + (2 x 1.01))g/mole = 12.0g/18.02g/mol = 0.665mole water

At first, it might appear that both will get used up, but, from the balanced equation we can see it takes 2 water molecules to react with each magnesium atom. So, water is the limiting reagant and some magnesium will be left over.

To determine to moles of excess Mg use the moles of reacants and the ratio from the balanced equation. Watch out, they want moles of Mg left over, not Mg used!

0.667mole mg - [0.665mole water x (1/2) ] =
0.335 mole Mg left over.

Answer 2

Agg! I have tried desperately. I do believe I have forgotten all of my highschool chemistry. And to think I wanted to major in the stuff.... I'm soo sorry! I really tried to help.

I know there's .67 moles of Mg and H2O... maybe that will help get you some where...
The balanced formula is 2Mg + 2H2O -> 2 MgH2 +O2... haha sorry I couldn't help more.

Answer 3

Mg(s) + 2H2O(l) --> Mg(OH)2(aq) + H2(g)

From the balanced equation, 1 mol Mg ~ 2 mol H2O.

Mol Mg = 16.2g x 1mol/24.3g = 0.667 mol Mg. We need
2 x 0.667mol of H2O = 1.33 mol H2O, so let's check it.

Mol H2O = 12.0 g x 1mol/18.0g = 0.667 mol H2O
We don't have enough H2O for 16.2 g of Mg, so H2O is the limiting reactant.

How much Mg is left after the reaction:
First, how much Mg is used?
mol Mg = 0.667 mol H2O x 1mol Mg/2 mol H2O
= 0.333 mol Mg used.

There were 0.667mol Mg available, 0.333mol Mg were used, so there is 0.334 mol Mg left.