Question

Under certain conditions, the substances sulfur dioxide and water combine to form sulfurous acid (H₂SO₃).

If 28.3 grams of sulfur dioxide and 8.0 grams of water combine to form sulfurous acid (H₂SO₃), how many grams of sulfurous acid (H₂SO₃) must form?

---------------grams sulfurous acid (H₂SO₃)

Answer 1

Molar mass of SO2,

MM = 1*MM(S) + 2*MM(O)

= 1*32.07 + 2*16.0

= 64.07 g/mol

mass(SO2)= 28.3 g

number of mol of SO2,

n = mass of SO2/molar mass of SO2

=(28.3 g)/(64.07 g/mol)

= 0.4417 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 8.0 g

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(8.0 g)/(18.016 g/mol)

= 0.444 mol

Balanced chemical equation is:

SO2 + H2O ---> H2SO3

1 mol of SO2 reacts with 1 mol of H2O

for 0.4417 mol of SO2, 0.4417 mol of H2O is required

But we have 0.444 mol of H2O

so, SO2 is limiting reagent

we will use SO2 in further calculation

Molar mass of H2SO3,

MM = 2*MM(H) + 1*MM(S) + 3*MM(O)

= 2*1.008 + 1*32.07 + 3*16.0

= 82.086 g/mol

According to balanced equation

mol of H2SO3 formed = (1/1)* moles of SO2

= (1/1)*0.4417

= 0.4417 mol

mass of H2SO3 = number of mol * molar mass

= 0.4417*82.09

= 36.3 g

Answer: 36.3 g