“Mole ratio is a conversion factor that proportionally relates the quantities of chemical substances worried in a response SI Unit: Mole
Mole Ratio = Moles of Reactants/Moles of products
consider the subsequent balanced chemical equation:
\({N_2} \left( g \right) + 3{H_2} \left( g \right) \rightarrow 2{NH_3} \left( g \right)\nonumber\)
The response has 6 mole ratios, elaborated as:
\(\begin{array}{ccc}\dfrac{3 \: \text{mol} \: {H_2}}{1 \: \text{mol} \: {N_2}} & or & \dfrac{1 \: \text{mol} \: {N_2}}{3 \: \text{mol} \: {H_2}} \\ \dfrac{1 \: \text{mol} \: {N_2}}{2 \: \text{mol} \: {NH_3}} & or & \dfrac{2 \: \text{mol} \: {NH_3}}{1 \: \text{mol} \: {N_2}} \\ \dfrac{3 \: \text{mol} \: {H_2}}{2 \: \text{mol} \: {NH_3}} & or & \dfrac{2 \: \text{mol} \: {NH_3}}{3 \: \text{mol} \: {H_2}} \end{array}\nonumber\)
Bear in mind you have got the following reaction:
\({N_2} \left( g \right) + 3{H_2} \left( g \right) \rightarrow 2{NH_3} \left( g \right)\nonumber\)
If 5.83 moles of hydrogen reacts with the extra amount of nitrogen, how many moles of ammonia will be produced after the reaction completes?
Answer:
Step No.1: Write the Given Values
Moles of \(H_2\) = 5.83
Step No.2: Now Write Down the Unknown amount
Moles of \(NH_3\) = ?
Step No.3: perceive the character of the problem
The given reaction is changing \(H_2\)into \(NH_3\) moles. So we are able to ignore the \(N_2\) completely. Our important concern is to discover a conversion issue that permits us to write down \(NH_3\) in the numerator and \(H_2\)
within the denominator.
Step No.4: solve
\(5.83 \: \text{mol} \:{H_2} \times \dfrac{2 \: \text{mol} \:{NH_3}}{3 \: \text{mol} \:{H_2}} = 3.88 \: \text{mol} \:{NH_3}\nonumber\)
