In a certain experiment 10.0g of magnesium nitride is allowed to react with 5.00 g of water. Calculate the final mass of ammonia, water, and magnesium nitride at the end of the reaction.

The final mass of ammonia is 1.57 grams

The final mass of water is 0 grams

The final mass of magnesium nitride is 5.34 grams

Explanation:

Step 1: Data given

Mass of Mg3N2 = 10.0 grams

Molar mass of Mg3N2 = 100.95 g/mol

Mass of H2O = 5.00 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

Step 3: Calculate moles Mg3N2

Moles Mg3N2 = mass / molar mass

Moles Mg3N2 = 10.0 grams / 100.95 g/mol

Moles Mg3N2 = 0.0991 moles

Step 4: Calculate moles H20

Moles H2O = 5.00 grams / 18.02 g/mol

Moles H2O = 0.277 moles

Step 5: Calculate the limiting reactant

H2O is the limiting reactant. It will completely be consumed (0.277 moles).

Mg3N2 is in excess. There will react 0.277/6 = 0.0462 moles

There will remain 0.0991 - 0.0462 = 0.0529 moles

This is 0.0529 moles * 100.95 g/mol = 5.34 grams

Step 6: Calculate moles of NH3

For 1 mol Mg3N2 we need 6 moles H2O to produce 3 moles Mg(OH)2 and 2 moles NH3

For 0.277 moles H2O we'll have 0.277/3 = 0.0923 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.0923 moles * 17.03 g/mol

Mass NH3 = 1.57 grams

you need to give more information

the answer for sodium is 0.186

the answer for rubidium is 0.251