Question

The mass percent of oxygen in 109 oleum is

Answer 1

109% oleum means we have now 109 gm of H2SO4.
Therfore, moles of H2SO4 = 109/98 = 1.112 moles
moles of O atoms = 1.112*4 = 4.448 moles.
weight = 4.448*16 = 71.18 gm.
% = 71.18/109 * 100
= 65.30 %

Answer 2

Answer: The percentage of oxygen in given amount of oleum is 65.3%.

Explanation: We are given 109% oleum and we need to calculate the mass percent of oxygen in it.

Oleum is the solution of $SO_3$ in $H_2SO_4$

As, 109% of oleum means we have 109 grams of $H_2SO_4$

Moles is calculated by using the formula:

$moles=\frac{\text{Given mass}}{\text{molar mass}}$

Molar mass of oleum = 98g/mol

Therefore, moles of $H_2SO_4=\frac{109}{98}=1.112moles$

Moles of O atoms = $1.112\times 4=4.448moles.$

Mass of oxygen = $4.448\times 16=71.18g$

Mass percentage of oxygen in oleum = $\frac{71.18}{109}\times 100=65.3\%$