Question

Calculate mass percent and mole percent of h2 and o2 if average molecular mass of the mixture is 14

Answer 1

Let in one mole of H2 and O2 mixture , x mol of H2 and (1 - x) mol of O2 are present.

so, average molar mass of mixture = mole of H2 × molar mass of H2 + mole of O2 × molar mass of O2

or, 14 = x × 2 + (1 - x) × 32

[ as we know, molar mass of H2 = 2g/mol and molar mass of O2 = 32g/mol]

or, 14 = 2x + 32 - 32x

or, -18 = -30x

or, x = 0.6 and (1 - x) = 0.4

so, number of mole of H2 = 0.6 mol

and number of mole of O2 = 0.4 mol

so, mole percent of H2 = 0.6/(0.6 + 0.4) × 100 = 60%

and mole percent of O2 = 100-60= 40%

again, mass of H2 in mixture = 0.6 × 2 = 1.2g

mass of O2 in mixture = 0.4 × 32 = 12.8g

mass percent of H2 = 1.2/(1.2 + 12.8) × 100 = 120/14 = 60/7 %

and mass percent of O2 = 100 - 60/7 = 640/7 %

Answer 2

Answer:

Both B and C are correct.

Explanation: