How many grams of methane need to react with 32 g of oxygen to make 22 g of $C O_{2}$ and 18 g of water?

32 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 8 g
Reaction:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
Moles of oxygen= $\frac{32}{32} = 1$
Moles of carbon dioxide= $\frac{22}{44} = 0.5$
Moles of water= $\frac{18}{18} = 1$
1 mole of methane reacts with 2 moles of oxygen to give 1 mole of $C O_{2}$ and 2 moles of water so 0.5 mole of methane will be required which is equal to $0.5 \times 16 = 8 g$ .

Suggest Corrections

Similar questions

22 g C O_{2}

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

22 g C O_{2} (g)

(O_{2})

0.20 g

(H_{2})

(H_{2} O)

H_{2} = 2 g/mol, O_{2} = 32 g/mol

0.2 g

Join BYJU'S Learning Program