Question

Calculate the weight of co2 and H2O that will be obtained by completely burning 0.25of an organic compound having molecular formula C4H4O4​

Answer 1

Answer : The weight of $CO_2$ and $H_2O$ will be, 0.3784 g and 0.0774 g respectively.

Solution :

Mass of $C_4H_4O_4$ = 0.25 g

Molar mass of $C_4H_4O_4$ = 116 g/mole

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_4H_4O_4$.

$\text{Moles of }C_4H_4O_4=\frac{\text{Mass of }C_4H_4O_4}{\text{Molar mass of }C_4H_4O_4}=\frac{0.25g}{116g/mole}=2.15\times 10^{-3}moles$

Now we have to calculate the moles of $CO_2$ and $H_2O$.

The given balanced chemical reaction is,

$C_4H_4O_4+3O_2\rightarrow 2H_2O+4CO_2$

From the balanced reaction, we conclude that

As, 1 mole of $C_4H_4O_4$ react to give 4 moles of $CO_2$

So, $2.15\times 10^{-3}moles$  moles of $C_4H_4O_4$ react to give $4\times (2.15\times 10^{-3})=8.6\times 10^{-3}$ moles of $CO_2$

As, 1 mole of $C_4H_4O_4$ react to give 2 moles of $H_2O$

So, $2.15\times 10^{-3}moles$  moles of $C_4H_4O_4$ react to give $2\times (2.15\times 10^{-3})=4.3\times 10^{-3}$ moles of $H_2O$

Now we have to calculate the mass of $CO_2$ and $H_2O$.

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(8.6\times 10^{-3}mole)\times (44g/mole)=0.3784g$

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

$\text{Mass of }H_2O=(4.3\times 10^{-3}mole)\times (18g/mole)=0.0774g$

Therefore, the weight of $CO_2$ and $H_2O$ will be, 0.3784 g and 0.0774 g respectively.