Chemistry question help

Question

For the following reaction, 163 grams of zinc oxide are allowed to react with 18.0 grams of water.

1. ZnO (s) + H₂O (l) Zn(OH)₂ (aq)

What is the FORMULA for the limiting reagent?

What is the maximum amount of zinc hydroxide that can be formed? 

What amount of the excess reagent remains after the reaction is complete? 

Answer 1

One easy way to find the limiting reagent is to divide the moles of each reactant by its coefficient in the balanced equation and whichever value is less is the limiting reagent.

ZnO + H₂O ==> Zn(OH)₂ ... balanced equation

For ZnO: 163 g ZnO x 1 mol ZnO/81.4 g = 2.00 mols ZnO (÷1->2.00)

For H₂O: 18.0 g H₂O x 1 mol H2O/18 g = 1.00 mols H2O (÷1->1.00)

Thus, H₂O is limiting since 1.00 is less than 2.00

To find out how much product is formed, use the moles of the limiting reagent:

1.00 mols H₂O x 1 mol Zn(OH)₂ / mol H₂O = 1.00 mols Zn(OH)₂ formed (if you want grams, multiply by molar mass of Zn(OH)₂

Amount of excess reagent:

Excess reagent is Zn, so find out how many moles of Zn is used, and subtract it from original moles of Zn.

moles of Zn used = 1.00 mols H₂O x 1 mol Zn / mol H₂O = 1.00 mols Zn used

moles of Zn left over= 2.00 mols - 1.00 mols = 1.00 mols Zn left over (if you want grams of Zn left over, multiple by the atomic mass for Zn).