You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

What is the v...

Question

What is the volume of water vapour required to produce 44.8 litres of oxygen by the reaction given below?

$2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$

100.0 l

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

89.6 l

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

44.8 l

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

22.4 l

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

11.2 l

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

Open in App

Solution

The correct option is D $89.6 l$
$2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$

From the above reaction, we can see that 1 L of oxygen is obtained from 2 L of water.

Therefore, 44.8 L of oxygen will be obtained from $2 \times 44.8 = 89.6$ L of water.

Hence, option B is correct.

Suggest Corrections

Similar questions

Q. What volume of

H_{2} O

is required to produce

5

O_{2}

by the following equation:

{2 H}_{2} O (g) \to {2 H}_{2} (g) + O_{2} (g)

Q. Given below is a balanced chemical reaction for the formation of ammonia.

N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g)

Estimate the volume of hydrogen gas required to produce 44.8 litres of

N H_{3}

. (Assume the reaction takes place at STP)

When 22.4 L of C $_{4}$ H $_{8}$ at STP is burnt completely, 89.6 L of CO $_{2}$ gas at STP and 72 g of water are produced. The volume of the oxygen gas at STP consumed in the reaction is closest to :

Q. When

22.4 L

C_{4} H_{8}

S T P

is burnt completely,

89.6 L

C O_{2}

gas the

S T P

and

72 g

of water are produced. The volume of the oxygen gas the STP consumed in the reaction is closest to

Q. Given below is a balanced chemical reaction for the formation of ammonia.

N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g)

Estimate the volume of hydrogen gas required to produce 44.8 litres of

N H_{3}

. (Assume the reaction takes place at STP)

Join BYJU'S Learning Program

What is the volume of water vapour required to produce 44.8 litres of oxygen by the reaction given below?2H2O(g)→2H2(g)+O2(g)

The correct option is D 89.6 l2H2O(g)→2H2(g)+O2(g)From the above reaction, we can see that 1 L of oxygen is obtained from 2 L of water.Therefore, 44.8 L of oxygen will be obtained from 2×44.8=89.6 L of water. Hence, option B is correct.

What is the volume of water vapour required to produce 44.8 litres of oxygen by the reaction given below?

$2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$

The correct option is D $89.6 l$
$2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g)$

From the above reaction, we can see that 1 L of oxygen is obtained from 2 L of water.

Therefore, 44.8 L of oxygen will be obtained from $2 \times 44.8 = 89.6$ L of water.

Hence, option B is correct.