Chemistry, 22.05.2020 02:06 Need1ng

When 100.g Mg3N2 reacts with 75.0 g H2O, 15.0 g of NH3 is formed. What is the % yield?

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Answer from: ksawyer745

The percentage yield of ammonia in the given reaction has been 44.44%.

The balanced chemical reaction can be:

$\rm Mg_3N_2\;+\;3\;H_2O\;\rightarrow\;3\;MgO\;+\;2\;NH_3$

The moles can be given as:

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Moles of $\rm Mg_3N_2$ = $\rm \dfrac{100}{100.95}$ mol

Moles of $\rm Mg_3N_2$ = 0.99 mol

Moles of $\rm H_2O$ = $\rm \dfrac{75}{18.02}$

Moles of $\rm H_2O$ = 4.16 mol

Moles of $\rm NH_3$ = $\rm \dfrac{15}{17.03}$

Moles of $\rm NH_3$ = 0.88 mol

The limiting reactant can be:

From the balanced equation for 3 moles of water, there has been 1 mole of magnesium nitrite. So, for 4.16 moles of water, there will be 1.38 moles of $\rm Mg_3N_2$ . Since the amount of $\rm Mg_3N_2$ has been completely consumed in the reaction, $\rm Mg_3N_2$ has been the limiting reactant.

So, the moles of ammonia produced =

1 moles $\rm Mg_3N_2$ = 2 moles ammonia.

0.99 moles $\rm Mg_3N_2$ = 0.99 $\times$ 2 moles $\rm NH_3$

0.99 moles $\rm Mg_3N_2$ = 1.98 moles of ammonia.

The theoretical yield of ammonia = 1.98 moles

The actual yield of ammonia = 0.88 mol.

% Yield of Ammonia = $\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100$

% Yield of Ammonia = $\rm \dfrac{0.88}{1.98}\;\times\;100$

% Yield of Ammonia = 44.44 %

For more information about percentage yield, refer to the link:

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Answer from: mfr81

% yield = 44.5 %

Explanation:

Step 1: Data given

Mass of Mg3N2 = 100.0 grams

Mass of H2O = 75.0 grams

Mass of NH3 = 15.0 grams

Molar mass of Mg3N2 = 100.95 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

Mg3N2 + 3H2O → 3MgO + 2NH3

Step 3: Calculate moles

Moles = mass / molar mass

Moles Mg3N2 = 100 grams / 100.95 g/mol

Moles Mg3N2 = 0.99 moles

Moles H2O = 75.0 grams / 18.02 g/mol

Moles H2O = 4.16 moles

Moles NH3 = 15.0 grams / 17.03 g/mol

Moles NH3 = 0.88 moles

Step 4: Calculate the limiting reactant

For 1 mol Mg3N2 we need 3 moles H2O to produce 3 moles MgO and 2 moles NH3

Mg3N2 is the limiting reactant. It will completely be consumed (0.99 moles)

H2O is in excess. There will react 3*0.99 = 2.97 moles

There will be produced 2*0.99 = 1.98 moles of NH3

Step 5: calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 1.98 moles * 17.03 g/mol

Mass of NH3 = 33.7 grams

Step 6: Calculate the % yield

%yield = (actual yield/ theoretical yield) * 100 %

% yield = (15.0 grams / 33.7) * 100 %

% yield = 44.5 %

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