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

Answer 2

A combustion reaction is a reaction in which the reactants burns in the presence of oxygen to yield the product.  The mass of water will be 17.3 grams.

What is mass?

Mass is the product of the molar mass and that of the number of moles of the substance. It is expressed in the terms of grams or kilograms.

Moles of benzene is calculated as:

$\begin{aligned} \rm moles &= \rm \dfrac{mass}{molar\; mass}\\\\&= \dfrac{25}{78}\\\\&= 0.32\;\rm mol\end{aligned}$

The combustion reaction can be shown as:

$\rm 2C_{6}H_{6} + 15O_{2} \rightarrow 12CO_{2} + 6H_{2}O$

From the reaction, it can be said that oxygen is the excess reagent and benzene is the limiting reagent.

If 2 moles of benzene = 6 moles of water

Then, 0.32 moles of benzene = 0.96 moles of water

From the mole, the mass of the water can be calculated as:

$\begin{aligned}\rm Mass &= \rm Molar\; mass\times moles\\\\&= 18 \times 0.96\\\\&= 17.3\;\rm gm\end{aligned}$

Therefore, the mass of water produced is 17.3 gm.

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