Question

in the following balanced equation 2C2H6 + 7O2 --> 4CO2 + 6H20 if 7.0 g of C2H6 react with 18g of O2, how many grams of water will be produced?

Answer 1

Answer: Amount of water produced is 8.6778 grams.

Explanation:

To calculate the amount of water produced, we first need to find the limiting reagent and for it we need to find the number of moles. To calculate the moles, we use the following equation:  

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

• Moles of $C_2H_6$:

Given mass of $C_2H_6$ = 7 grams

Molar mass of $C_2H_6$ = 30 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{7g}{30g/mol}=0.2333moles$

• Moles of $O_2$:

Given mass of $O_2$ = 18 grams

Molar mass of $O_2$ = 32 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{18g}{32g/mol}=0.5625moles$

For the given chemical reaction, the equation follows:

$2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O$

By Stoichiometry:

7 moles of oxygen reacts with 2 moles of ethane.

So, 0.0261 moles of lead nitrate are produced by = $\frac{2}{7}\times 0.5625=0.1607moles$ of ethane.

As, the required amount of ethane is less than the given amount. Hence, it is considered as the excess reagent.

Oxygen gas is considered as a limiting reagent because it limits the formation of product.  

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 6 moles of water.

So, 0.5625 moles of oxygen gas will produce = $\frac{6}{7}\times 0.5625=0.4821moles$ of water.

• Now, to calculate the mass of water produced, we use equation 1.

Molar mass of water = 18 g/mol

Moles of water produced = 0.4821 moles

Putting values in the equation, we get:

$0.4821mol=\frac{\text{Mass of water}}{18g/mol}$

Mass of water produced = 8.6778 grams.

Hence, amount of water produced is 8.6778 grams.