23. In the following reaction, how many grams of sodium sulfate are needed to react with...

23. In the following reaction, how many grams of sodium sulfate are needed to react with 3.08 moles of strontium nitrate? (Hint: stoichiometry mol Na2SO4: mol Sr(NO3)2 = 1 mol: 1 mol) Sr(NO3)2 (aq) + Na2SO4 (aq) --> SrSO4 (s) + 2 NaNO3 (aq) (balanced?)

Expert Solution

Sr(NO₃)₂ (aq) + Na₂SO₄ (aq) SrSO₄ (s) + 2 NaNO₃ (aq)

Molar mass of Na₂SO₄ = 142.04 g/mol

Molar mass of Sr(NO₃)₂ = 211.63 g/mol

In this reaction,

1 mole of Na₂SO₄ reacts with 1 mole of Sr(NO₃)₂.

3.08 moles of Na₂SO₄ reacts with 3.08 moles of Sr(NO₃)₂.

Now, 1 mole of Na₂SO₄ = 142.04 g

3.08 moles of Na₂SO₄ = (3.08 x 142.04) g = 437.48 g

So, 437.48 g of sodium sulfate are needed to react with 3.08 moles of strontium nitrate.

queen_honey_blossom answered 2 years ago

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