How many moles of $K C l O_{3}$ would be required to supply the proper amount of oxygen needed to burn $24$ g of $C H_{4}$ completely?
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$

Open in App

Solution

The balanced chemical equation for the complete consumption of methane is as given below:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$24$ g of methane corresponds to $\frac{24 g}{16 g / m o l} = 1.5$ mol.
They will require $\frac{2}{1} \times 1.5 m o l = 3.0$ mol.
Thus, $3$ moles of oxygen are required to react completely with $24$ g of methane.
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
Thus, $3$ moles of oxygen will be supplied by $2$ moles of $K C l O_{3}$ .

Suggest Corrections

Similar questions

24

C H_{4}

C H_{4}

425.6 k c a l

C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O + 212.8 k c a l

O_{2}

22.4 d m^{3}

C H_{4}

11.2 d m^{3}

H_{2}

C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O

2 H_{2} + O_{2} \to 2 H_{2} O

O_{2}

10 c m^{3}

C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O

2 C_{2} H_{2} + 5 O_{2} ⟶ 4 C O_{2} + 2 H_{2} O

Join BYJU'S Learning Program