Percent Yeild

When reacting Na with Cl₂, we calculated that the theoretical yield should be 12.5 grams. Our actual yield was 13.0 grams. What is the percent yield?

• 100%
• 104%
• 96%
• 1.04

Answer explanation

13.0 g / 12.5 x 100

actual/theoritical x 100

2Fe₂O₃  +  C  →  Fe  +  3CO₂
You add 28 grams of carbon. You find the actual yield to be 181.2 grams of CO₂. What is the percent yield of CO₂? (Hint: calculate theoretical yield of CO₂ first)

• 158.9%

• 209.2%
• 6435%
• 15.5

Answer explanation

2Fe₂O₃  +  C  →  Fe  +  3CO₂
GIVEN: 28 grams of carbon.

WANTED: grams CO₂

actual / theoritical x 100

Actual = 181.2 grams of CO₂

Theoritical = 307. 7

Theoretical yield = 73g
Actual yield = 62g
Calculate the percent yield.

• 1.16%
• 116%
• 85%
• 76%

Answer explanation

Actual yield / Theoritical x 100 = percent yeild

62g / 73g x 100 =

In a lab, a scientist calculate he should produce 12.3 grams of product in his experiment. When he is finished collecting his product it weighs 10.1 grams. What is his percent yield?

• 82%
• 0.82%
• 10.1 %
• 100%

CH₄  +  2O₂  →  CO₂  +  2H₂O
24 grams of CH₄ was added to the above reaction. Calculate the theoretical yield of CO₂.

• 68 grams

• 132 grams
• 33 grams
• 8.72

Answer explanation

CH₄  +  2O₂  →  CO₂  +  2H₂O
GIVEN: 24 grams of CH₄

WANTED: g CO₂ (theoritical yeild)

2Fe₂O₃ + C → Fe + 3CO₂

You add 28 grams of carbon. You find the actual yield to be 181.2 grams of CO₂. What is the percent yield of CO₂? (Hint: calculate theoretical yield of CO₂ first)

• 58.83%

• 308%

• 6435%

• 15.5

The reaction of 25.0 g benzene, C₆H₆, with excess HNO₃ resulted in 21.4 g C₆H₅NO₂. What is the percentage yield?

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

• 100%

• 27.39%

• 54.29%

• 85.62%

Answer explanation

GIVEN: 25.0 g benzene,

WANTED: g C₆H₅NO₂ (theoritical)

with excess HNO₃ = plenty of HNO₃

Actual 21.4 g C₆H₅NO₂.

Lead nitrate can be decomposed by heating. What is the percent yield of the decomposition reaction if 7.4 g Pb(NO₃)₂ are heated to give 4.1 g PbO?

2 Pb(NO₃)_{2 (s)}→2 PbO_(s)+ 4 NO_2(g)+ O_2(g)

• 82%

• 44%

• 56%

• 67%