Answer: 1.848 g
Explanation: To calculate the moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}\) ....(1)
For ammonia:
Putting values in above equation, we get:
\(\text{Moles of ammonia}=\frac{8.00\times 10^{22}}{6.023\times 10^{23}}=0.132mol\)
\(\text{Moles of oxygen}=\frac{7.00times 10^{22}}{6.023\times 10^{23}}=0.116mol\)
For the reaction:
\(4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)\)
By Stoichiometry of the reaction,
4 moles of ammonia combine with 3 moles of Oxygen
Thus 0.132 moles of ammonia will combine with=\(\frac{3}{4}\times 0.132=0.009mol\) of oxygen
Thus ammonia is the limiting reagent as it limits the formation of product.
4 moles of ammonia produces 2 moles of nitrogen
0.132 moles of ammonia will produce=\(\frac{2}{4}\times 0.132=0.066 moles\) of nitrogen
Molar mass of nitrogen = 28 g/mol
Amount of nitrogen produced=\({text {no of moles}}\times {text {molar mass}}=0.066\times 28=1.848g\)
