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.
