Question

volume of oxygen gas at stp in litres required to burn 1.6g of methane​

Answer 1

I took 4g please u take 1.6 g

Answer 2

At STP, 4.48 L of oxygen gas is required to burn 1.6g of methane.

• The chemical equation involved in the combustion of methane is as follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

• From the above equation, it is inferred that, for complete combustion of 1 mole of methane ($CH_4$), 2 moles of oxygen are required.
• To find number of moles, the following calculation must be done:

Molar mass of methane = 12 + 4(1) = 16 g

No. of moles of 1.6g of methane = $\frac{given mass}{Molar mass}=$ $\frac{1.6}{16}= 0.1$ mole

We know that,

2 mole of $O_2$ is needed for complete combustion of 1 mole of methane.

∴ For 0.1 mole of methane = 0.1 ×2 = 0.2 mole

At STP, 1 mole of all gas has a volume of 22.4L

So, 0.2 mole of oxygen gas will require = 22.4 × 0.2 = 4.48 L

• Thus it is inferred that at STP, 4.48 L of oxygen gas is required to burn 1.6g of methane.

Learn more about such concept

At STP 2 g of helium gas (molar mass = 4) occupies a volume of

(1) 22.4 dm (2) 11.2 dm (3) 5.6 dm (4) 2 dm​

brainly.in/question/12153386

From a container having 64g oxygen ,11.2 l oxygen gas at s.t.p and 6.022 *10^23 oxygen atoms are removed. find the mass of the oxygen gas left

brainly.in/question/3731820