50.0 kg of N2 and 10.0 kg of H2 are mixed to produce NH3.calculate the NH3 fromed.identify the limiting reagent in the production of NH3 in this situation.
Answers
Answer:
I can't understand your question
Answer:
56.1 kg of Ammonia is formed. (The answer may be long but I have explained it step by step.)
Explanation:
The equation goes like this:
N2 + 3H2 -----> 2 NH3
Calculation of moles:
No. of moles = mass/ molecular mass
No. of moles of N2 = 50000/28 =1786.7 = 17.86 x 10^2 moles
No. of moles of H2 = 10000/2 = 5000 = 50 x 10^2 moles
Finding limiting reagent:
From the above equation:
1 mole of N2 needs 2 moles of H2 to complete the reaction
So 17.86 x 10^2 moles needs x moles of hydrogen
cross multiply:
x = 3 x (17.86 x 10^2) moles
= 53.58 x 10^2 moles
But we have only 50 x 10^2 moles. So hydrogen is the limiting reagent.
Shortcut method to find limiting reagent:
The substance which is less amount is the limiting reagent, because it gets consumed first fully.
Finding the no. of moles of NH3 formed:
3 moles of H2 gives 2 moles of NH3
So, 50 x 10^2 moles of H2 will give x moles of NH3
CROSS MULTIPLY:
x*3 = 2 * 50 * 10^2
x = 100 * 10^2 / 3
x = 10 * 10^3 / 3
x = 3.3 * 10^3
Converting moles into mass:
no. of moles = mass / molecular mass
mass= no. of moles x molecular mass
molecular mass of NH3= 14 + 3 =17g
mass = 3.3 x 10^3 x 17
= 56.1 x 10^3 g
= 56.1 kg
( Since prefix of kg is 10^3)