Question

Consider the following Reaction.

CH4 (g) + O2 (g) CO2 (g) + H2O (l)

A chemist allows 23.2g of CH4 and 78.3g O2 to react. When the reaction is
finished, the chemist collects 52.7g CO2. Determine the limiting reagent,
theoretical yield, and percent yield for the reaction

Answer 1

Answer: -

O 2 limiting reagent

53.83 g CO2 theoretical Yield

97.9% percentage yield

Explanation: -

Mas of CH4 = 23.2 g

Molar mass of CH4 = 12 x 1 + 1 x 4 = 16g

Mass of O2 = 78.3 g

Molar mass of O 2 – 16 x 2 = 32 g

The balanced chemical equation for the reaction is

CH4 + 2 O2 = CO2 + 2 H2O

From the balanced equation we see that

2 O 2 reacts with 1 CH4

2 x 32 g of O 2 react with 16 g of CH4

78.3 g of O 2 react with
`(16 g CH4 x 78.3 g O2)/(2 x32 g O2)`

= 19.575 g pf CH4

Thus CH4 is in excess.

The limiting reagent is thus O 2.

Molar mass of CO2 = 12 x 1 + 16 x 2= 12 +32 = 44g

From the balanced equation we see

2O 2 gives 1 CO2

2 x 32 g of O 2 gives 44g of CO2

78.3 g of O 2 gives =
`(44 gCO2 x 78.3 g O2)/(2 x 32 g O2)`

=53.83 g of CO2

Theoritical Yield = 53.83 g of CO2

Actual yield = 52.7 g

Percentage yield =
`(52.7 g)/(53.83 g) x 100`

=97.9 %