Question

​SO2 + 2NaOH → Na2SO3 + H2O

Answer 1

Answer:

There will be produced 157.55 grams of Na2SO3. The limiting reactant is NaOH. SO2 is in excess, there will remain 19.9 grams of SO2.

Explanation:

Step 1: Data given

Mass of SO2 = 100 grams

Mass of NaOH = 100 grams

Molar mass of SO2 = 64.07 g/mol

Molar mass of NaOH = 40 g/mol

Molar mass of Na2SO3 = 126.04 g/mol

Step 2: The balanced equation

​SO2 + 2NaOH → Na2SO3 + H2O

Step 3: Calculate moles SO2

Moles SO2 = mass SO2 / molar mass SO2

Moles SO2 = 100.0 grams / 64.07 g/mol

Moles SO2 = 1.561 grams

Step 4: Calculate moles of NaOH

Moles NaOH = 100.0 grams / 40 g/mol

Moles NaOH = 2.5 moles

Step 5: Calculate limiting reactant

For 1 mol of SO2 we need 2 moles of NaOH to produce 1 mol Na2SO3 and 1 mol of H2O

NaOH is the limiting reactant. It will completely be consumed.(2.5 moles).

SO2 is in excess. There will be consumed 2.5 / 2 = 1.25 moles of SO2

There will remain 1.561 - 1.25 = 0.311 moles of SO2. This is 0.311 * 64.07 g/mol = 19.9 grams

Step 6: Calculate moles of Na2SO3

There will be produced 1.25 moles of Na2SO3

Step 7: Calculate mass of Na2SO3

Mass Na2SO3 = 1.25 * 126.04 g/mol

Mass Na2SO3 = 157.55 grams

There will be produced 157.55 grams of Na2SO3. The limiting reactant is NaOH. SO2 is in excess, there will remain 19.9 grams of SO2.