Question

Using the following equation

Answer 1

Thermal energy when solid, liquor , and gas combine

Answer 2

Answer : The mass of $H_2O$ produced will be, 17 grams.

Explanation : Given,

Mass of $C_2H_6$ = 9.5 g

Mass of $O_2$ = 130 g

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_2H_6$ and $O_2$.

$\text{Moles of }C_2H_6=\frac{\text{Mass of }C_2H_6}{\text{Molar mass of }C_2H_6}=\frac{9.5g}{30g/mole}=0.317moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{130g}{32g/mole}=4.06moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 2 moles of $C_2H_6$ react with 7 mole of $O_2$

So, 0.317 moles of $C_2H_6$ react with $\frac{7}{2}\times 0.317=1.109$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_6$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$.

As, 2 moles of $C_2H_6$ react to give 6 moles of $H_2O$

So, 0.317 moles of $C_2H_6$ react to give $\frac{6}{2}\times 0.317=0.951$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(0.951mole)\times (18g/mole)=17.118g\approx 17g$

Therefore, the mass of $H_2O$ produced will be, 17 grams.