Question

Given the following equation: > 8 CO2 10 H20 2 С4Н10 + 13 02 How many grams of C4H10 are needed to react with 35.1 grams of O2?

Answer 1

Answer: The mass of butane reacting with oxygen gas is 9.76 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    .....(1)

• For oxygen

Given mass of oxygen gas = 35.1 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{35.1g}{32g/mol}=1.096mol$

For the given chemical reaction:

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

By stoichiometry of the reaction:

13 moles of oxygen gas is reacting with 2 moles of butane.

So, 1.096 moles of oxygen gas will react with = $\frac{2}{13}\times 1.096=0.168moles$ of butane.

Now, calculating the mass of butane from equation 1, we get:

Molar mass of butane = 58.12 g/mol

Moles of butane = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of butane}}{58.12g/mol}\\\\\text{Mass of butane}=9.76g$

Hence, the mass of butane reacting with oxygen gas is 9.76 grams.