Question

Consider the following 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 $\frac{16 g CH4 x 78.3 g O2}{2 x32 g O2}$

= 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 = $\frac{44 gCO2 x 78.3 g O2}{2 x 32 g O2}$

=53.83 g of CO2

Theoritical Yield = 53.83 g of CO2

Actual yield = 52.7 g

Percentage yield = $\frac{52.7 g}{53.83 g} x 100$

=97.9 %