Respuesta :
Answer:
% 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 %
The percentage yield of ammonia in the given reaction has been 44.44%.
The balanced chemical reaction can be:
[tex]\rm Mg_3N_2\;+\;3\;H_2O\;\rightarrow\;3\;MgO\;+\;2\;NH_3[/tex]
The moles can be given as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
- Moles of [tex]\rm Mg_3N_2[/tex] = [tex]\rm \dfrac{100}{100.95}[/tex] mol
Moles of [tex]\rm Mg_3N_2[/tex] = 0.99 mol
- Moles of [tex]\rm H_2O[/tex] = [tex]\rm \dfrac{75}{18.02}[/tex]
Moles of [tex]\rm H_2O[/tex] = 4.16 mol
- Moles of [tex]\rm NH_3[/tex] = [tex]\rm \dfrac{15}{17.03}[/tex]
Moles of [tex]\rm NH_3[/tex] = 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 [tex]\rm Mg_3N_2[/tex]. Since the amount of [tex]\rm Mg_3N_2[/tex] has been completely consumed in the reaction, [tex]\rm Mg_3N_2[/tex] has been the limiting reactant.
So, the moles of ammonia produced =
1 moles [tex]\rm Mg_3N_2[/tex] = 2 moles ammonia.
0.99 moles [tex]\rm Mg_3N_2[/tex] = 0.99 [tex]\times[/tex] 2 moles [tex]\rm NH_3[/tex]
0.99 moles [tex]\rm Mg_3N_2[/tex] = 1.98 moles of ammonia.
The theoretical yield of ammonia = 1.98 moles
The actual yield of ammonia = 0.88 mol.
% Yield of Ammonia = [tex]\rm \dfrac{Actual\;yield}{Theoretical\;yield}\;\times\;100[/tex]
% Yield of Ammonia = [tex]\rm \dfrac{0.88}{1.98}\;\times\;100[/tex]
% Yield of Ammonia = 44.44 %
For more information about percentage yield, refer to the link:
https://brainly.com/question/12809634
