Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 8.83 g of sulfuric acid is mixed with 9.1 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.

13 grams Na₂SO₄ (2 sig.figs.)

Explanation:

1st convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.

Determine limiting reactant and

Complete problem by converting yield into grams.

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄

moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH

Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.

H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)

NaOH =>  (0.2275/2) = 0.1138

NOTE: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...

                    H₂SO₄        +        2NaOH       =>       Na₂SO₄      +      2H₂O

moles      0.0901 mole          0.2275 mol            0.0901 mol      2(0.0901 mol)

mass (g) Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 12.798 grams ≅ 13 grams                                    Na₂SO₄ (2 sig.figs.)

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"