Question

Percent Yield Supplemental Practice Problems
1. What is the percent yield for the following reaction given we start with 34.08 grams of ammonia and 96.00 grams of oxygen and obtain an actual yield of 21.73 grams of nitric oxide?
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

2. What is the percent yield for the following reaction given we start with 32.38 grams of aluminum and 609.12 grams of iodine and obtain an actual yield of 427.92 grams of aluminum iodide?
2 Al(s) + 3 I2(s) → 2 AlI3(s)


3. What is the percent yield for the following reaction given we start with 50.6 grams of magnesium hydroxide and 45.0 grams of hydrochloric acid and obtain an actual yield of 51.7 grams of magnesium chloride?
Mg(OH)2(aq) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l)

4. What is the percent yield for the following reaction given we start with 34.2 grams of ethane (C2H6) and 62.7 grams of oxygen and obtain an actual yield of 24.1 grams of water?
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

Answer 1

1)

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 34.08 g

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(34.08 g)/(17.034 g/mol)

= 2.001 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 96.0 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(96.0 g)/(32 g/mol)

= 3 mol

Balanced chemical equation is:

4 NH3 + 5 O2 ---> 4 NO + 6 H2O

4 mol of NH3 reacts with 5 mol of O2

for 2.000704 mol of NH3, 2.500881 mol of O2 is required

But we have 3 mol of O2

so, NH3 is limiting reagent

we will use NH3 in further calculation

Molar mass of NO,

MM = 1*MM(N) + 1*MM(O)

= 1*14.01 + 1*16.0

= 30.01 g/mol

According to balanced equation

mol of NO formed = (4/4)* moles of NH3

= (4/4)*2.000704

= 2.001 mol

mass of NO = number of mol * molar mass

= 2.001*30.01

= 60.04 g

% yield = actual mass*100/theoretical mass

= 21.73*100/60.04

= 36.2 %

Answer: 36.2 %

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