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

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2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !