When solid magnesium nitride reacts with water the reaction produces solid magnesium hydroxide and ammonia gas. 26.0 grams of magnesium nitride are reacted with 17.6 grams of water.

Question

1. How many grams of magnesium nitride will remain after reaction is complete?
2. How many grams of water will reamain after reaction is complete?
3. How many grams of magnesium hydroxide will remain after reaction is complete?
4. How many grams of ammonia gas will remain after reaction is complete?

Answer 1

This is a limiting reagent (LR) problem.
Mg3N2 + 6HOH --> 3Mg(OH)2 + 2NH3

mols Mg3N2 = g/molar mass = ?
mols H2O = g/molar mass = ?
Using the coefficients in the balanced equation, convert mols Mg3N2 to mols Mg(OH)2
Do the same to convert mols H2O to mols Mg(OH)2. It is likely that you will get two answers for mols Mg(OH)2 formed. In LR problems, the SMALLER number of mols produced is correct and the reagent providing that number (in this case either Mg3N2 or H2O). The other reagent we will call OR.
3. g Mg(OH)3 formed (remaining)= mols Mg(OH)2 x molar mass = ?
4. Using the coefficients in the balanced equatioin, convert mols of the LR to mols NH3. Then g NH3 formed (remaining) = mols NH3 x molar mass NH3 = ?
1 and 2. The LR will have zero mols remaining and zero grams remaining because all of it will be used. That's why it is called the LR. To find the other one, Convert mols LR to mols OR used. Suibtract from mols OR initially to find mols OR remaining. Then grams = mols x molar mass.
Post your work if you get stuck.