Question

Consider the reaction for the combustion of benzene: 2 C6H6(g) 15 O2(g) ---> 12 CO2(g) 6 H2O(l) What mass of H2O is produced when 25.0 g of benzene reacts with excess O2(g) (i.e., all benzene is consumed)

Answer 1

Answer: Mass of water produced is 17.3 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 benzene:

Given mass of benzene = 25.0 g

Molar mass of benzene= 78 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzene}=\frac{25.0g}{78g/mol}=0.32mol$

The chemical equation for the reaction follows:

$2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$

As, oxygen is an excess reagent. Thus benzene is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of benzene produces 6 moles of water

So, 0.32 moles of benzene will produce = $\frac{6}{2}\times 0.32=0.96moles$ of water

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

Molar mass of water= 18 g/mol

Moles of water = 0.96moles

Putting values in equation 1, we get:

$0.96mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=17.3g$

Thus mass of water produced is 17.3 grams