Question

Under certain conditions, the substances sulfur dioxide and water combine to form sulfurous acid (H_2_SO_3_).If 26.2 grams of sulfur dioxide and 7.4 grams of water combine to form sulfurous acid (H_2_SO_3_), how many grams of sulfurous acid (H_2_SO_3_) must form?

Answer 1

Step-by-step explanation

Given

Mass of SO2 = 26.2 g

Mass of H2O = 7.4 g

Required: Mass of H2SO3

Solution

Step 1: write a balanced chemical equation

`SO_2+H_2O\rightarrow H_2SO_3`

Step 2: Calculate the moles of H2SO3 to be produced by SO2

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 26.2g/64,066 g/mol

n = 0.4089 mol

molar ratio between SO2 and H2SO3 is 1:1 there moles of H2SO3 = 0.4089 mol

Step 3: Calculate the moles of H2SO3 to be produced by H2O

n = m/M where n is the moles, m is the mass and M is the molar mass

n = 7.4g/18,01528 g/mol

n = 0.4108 mol

molar ratio between H2O and H2SO3 is 1:1 there moles of H2SO3 = 0.4108 mol

Therefore the limiting reagent is SO2 because it has less number of moles.

Step 4: Calculate the mass of H2SO3

m = n x M

m = 0.4089 x 82,07 g/mol

m = 33.56 g

Answer

mass of H2SO3 = 33.56 g