Question

Using the following equation

2C2H6 +7O2 -->4CO2 +6H2O

if 9.5g C2H6 react with 130g of O2, how many grams of water will be produced?

Question options:

32


41


9


17

Answer 1

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}=(9.5g)/(30g/mole)=0.317moles`

`\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=(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
`(7)/(2)* 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
`(6)/(2)* 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* \text{Molar mass of }H_2O`

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

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