Question

​SO2 + 2NaOH → Na2SO3 + H2O

There are 100.0 grams of each reactant available. Determine the limiting reactant in this equation.

Answer 1

To find the limiting reactant when 100.0 grams of SO2 and 100.0 grams of NaOH are available, we first calculate the number of moles of each reactant. Upon comparing the stoichiometry of the balanced equation with the available moles, NaOH is determined to be the limiting reactant because there are fewer moles available than required to react with all of the SO2.

Step-by-step explanation:

To determine the limiting reactant in the reaction SO2 + 2NaOH → Na2SO3 + H2O, where 100.0 grams of each reactant is available, we need to start by calculating the number of moles of each reactant.

First, we determine the molar masses of SO2 and NaOH. The molar mass of SO2 (32.07 g/mol for S + 2 × 16.00 g/mol for O) is approximately 64.07 g/mol. The molar mass of NaOH (22.99 g/mol for Na + 15.999 g/mol for O + 1.008 g/mol for H) is roughly 40.00 g/mol.

Now, we divide the mass of each reactant by its respective molar mass to get the moles:

• For SO2: 100.0 g ÷ 64.07 g/mol = 1.56 moles
• For NaOH: 100.0 g ÷ 40.00 g/mol = 2.50 moles

Looking at the stoichiometry of the balanced chemical equation, we see that 1 mole of SO2 reacts with 2 moles of NaOH. Thus, for 1.56 moles of SO2, we would need 3.12 moles of NaOH, but we have only 2.50 moles available. Therefore, NaOH is the limiting reactant because we do not have enough NaOH to react with all of the SO2.

Answer 2

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.

Step-by-step 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.