so2 + 2naoh → na2so3 + h2o

there are 100.0 grams of each reactant available. determine the limiting reactant in this equation.

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.