Using the following balanced equation: 2 NaOH (aq) + H2SO4 (aq) → 2 H2O (l) + Na2SO4 (aq). How many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid?

step 1: number of moles (n) = mass (m) divided by molecular mass (Mm)

Mm of NaOH = 40 g/mole

n of NaOH = 200/40 = 5 moles

step 2:

from the chemical reaction provided: 2NaOH + H2SO4 → 2H2O + Na2SO4

2 moles of NaOH produce 1 mole of Na2SO4 (sodium sulfate)

n of Na2SO4 = 5 moles NaOH x (1 mole Na2SO4/2 moles NaOH) = 2.5 moles Na2SO4

step 3:

n=m/Mm thus mass (m) = n x Mm

Mm of Na2SO4 = 142.04 g/mole

m = 2.5 x 142.02 = 355.1 grams of Na2SO4 will be produced

Explanation:

In step 1: we prefer to work with number of moles therefore convert the mass into moles

In step 2: With the chemical reaction, use the mole ratio 2:1 to find the number of moles of Sodium sulfate from the calculated NaOH

In step 3: with Na2SO4 moles calculated, find the mass using the formula

and the mass is calculated to be 355.1 grams