If 116 g of butane burns in the presence of 320. g of oxygen how much carbon dioxide will be produced? CO2 produced in grams

270.77 g of CO2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of C4H10 and O2 that reacted and the mass of CO2 produced from the balanced equation. This can be obtained as follow:

Molar mass of C4H10 = (12×4) + (10×1)

= 48 + 10 = 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 2 × 16 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of C4H10 reacted with 416 g of O2 to produce 352 g of CO2.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of C4H10 reacted with 416 g of O2.

From the above, we can see that a higher mass of O2 i.e 416 g than what was given i.e 320 g his needed to react completely with 116 g of O2. Therefore, O2 is the limiting reactant and C4H10 is the excess reactant.

Finally, we shall determine the mass of CO2 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of CO2.

The limiting reactant is O2 and the mass of CO2 produced can be obtained as follow:

From the balanced equation above,

416 g of O2 reacted to produce 352 g of CO2.

Therefore, 320 g of O2 will react to produce = (320 × 352)/416 = 270.77 g of CO2.

Thus, 270.77 g of CO2 were obtained from the reaction.

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