Question

For items 8 to 11. In the reaction 2H2 + CO → CH3OH, 6.8 g carbon monoxide gas was made to react with 7.2 g of hydrogen gas. The reaction produced 5.2 g of methanol.

8. Which is the limiting reagent?
A. Hydrogen
B. Carbon monoxide
C. methanol
D. CO2
iv

9. What is the theoretical yield?
A. 57.6 g
B. 50.4 g
C. 7.8 g
D. 6.8 g

10. What is the percent yield?
A. 76%
B. 78%
C. 68%
D. 67%

Answer 1

Answer:8.D.CO2

9.B.50.4

10.c.68%

Explanation:hope it   is right

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Answer 2

                             2H₂ + CO → CH₃OH

Given:

• 6.8 g of CO
• 7.2 g of H₂
• 5.2 g of CH₃OH

1. The limiting reagent is CO.

• Limiting reagent is the one which produces lesser amount of the product.
• In the given reaction,
• 1 mole of CO produces 1 mole of CH₃OH.

     = 28 g of CO produces  = 32 g of CH₃OH.

    = 1 g of CO produces  = 32/28 g of CH₃OH.

    = 6.8 g CO produces  = (32/28) x 6.8 g of CH₃OH.

                                        =  7.771 g

• 2mole H₂ of produces 1 mole of CH₃OH.

    =4g H₂ of produces  = 32 g of CH₃OH.

    =1g H₂ of produces  = 32/4 g of CH₃OH.

    =7.2g H₂ of produces  = (32/4) x 7.2 g of CH₃OH.

                                          = 57.6 g

• As CO produces less amount of methanol, therefore CO is the limiting reagent.

2. The theoretical yield is 7.8 g.

• From the above calculations,
• 6.8 g CO produces  = 7.771 g ≈ 7.8g

3. The percentage yield is 66.7%

      % yield = (actual yield/theoretical yield) x 100

                   = 5.2/7.8 x 100

                   = 66.6% ≈ 66.7%