Question

SO3 (g) + H2O (l) ------> H2SO4 (l)

Answer 1

Answer:

Moles of $H_2O$ reacted = 0.1529 moles

Explanation:

Mass of $H_2SO_4$ obtained = 15 g

Thus, formation of the product can be used to determine the moles of water reacted as:-

Molar mass of $H_2SO_4$ = 98.079 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{15\ g}{98.079\ g/mol}$

$Moles_{H_2SO_4}= 0.1529\ mol$

According to the given reaction:-

$SO_3+H_2O\rightarrow H_2SO_4$

1 mole of $H_2SO_4$ forms when 1 mole of $H_2O$ is reacted

So,

0.1529 mole of $H_2SO_4$ forms when 0.1529 mole of $H_2O$ is reacted

Thus, Moles of $H_2O$ reacted = 0.1529 moles