Suppose 6.87 g of sulfuric acid is mixed with 9.7 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

9.96g of Na₂SO₄ is the maximum mass that could be produced

Explanation:

We must determine the reaction:

Reactants: H₂SO₄, NaOH

Products: H₂O, Na₂SO₄

The equation is:  H₂SO₄ (aq) + 2NaOH(aq) →  2H₂O (l) + Na₂SO₄(aq)

We have the mass of the reactants. We need to convert them to moles, in order to define the limiting reactant

6.87 g / 98 g/mol = 0.0701 moles of acid

9.7 g / 40 g/mol = 0.242 moles of base

Limiting reactant is the acid. Let's verify

2 moles of NaOH can react with 1 mol of acid

Therefore 0.242 moles of NaOH must react with (0.242 . 1) / 2 = 0.121 moles

We do not have enough acid.

Ratio with the salt is 1:1. 1 mol of acid produces 1 mol of salt

Therefore 0.0701 moles of acid will produce 0.0701 moles of salt

We convert the moles to mass → 0.0701 mol . 142.06 g / 1 mol = 9.96g

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying