2.00 g Ho is allowed to react with 10.5 g Na, producing 2 48 g NH ndustnal proce

Question

2.00 g Ho is allowed to react with 10.5 g Na, producing 2 48 g NH ndustnal process In the Haber-Bosch process hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units 3H2(a) + N2(8)->2NHs(g) produced in the Haber-Bosch process has a wide range of uses, from fertilizer to Hints of ammonia is difficult resulting in lower yields than those predicted from the chemical equation. Part B Express your answer to three significant figures Hints PM

Explanation / Answer

A)

Molar mass of N2 = 28.02 g/mol


mass(N2)= 10.5 g

number of mol of N2,
n = mass of N2/molar mass of N2
=(10.5 g)/(28.02 g/mol)
= 0.3747 mol

Molar mass of H2 = 2.016 g/mol


mass(H2)= 2.0 g

number of mol of H2,
n = mass of H2/molar mass of H2
=(2.0 g)/(2.016 g/mol)
= 0.9921 mol
Balanced chemical equation is:
N2 + 3 H2 ---> 2 NH3 +


1 mol of N2 reacts with 3 mol of H2
for 0.3747 mol of N2, 1.1242 mol of H2 is required
But we have 0.9921 mol of H2

so, H2 is limiting reagent
we will use H2 in further calculation


Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol

According to balanced equation
mol of NH3 formed = (2/3)* moles of H2
= (2/3)*0.9921
= 0.6614 mol


mass of NH3 = number of mol * molar mass
= 0.6614*17.03
= 11.27 g
Answer: 11.3 g

B)

% yield = actual mass*100/theoretical mass
= 2.48*100/11.27
= 22.0 %
Answer: 22.0 %

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