Question

If 90.00 grams of calcium nitrate are used up in the reaction how many grams of aluminum sulfate are necessary to react ?

Answer 1

Answer:

62.56 g

Explanation:

The reaction between calcium nitrate, Ca(NO₃)₂, and aluminum sulfate, Al₂(SO₄)₃, is:

3Ca(NO₃)₂ + Al₂(SO₄)₃ → 3CaSO₄ + 2Al(NO₃)₃

The molar masses are:

Ca(NO₃)₂ = 40 g/mol of Ca + 2*14 g/mol of N + 6*16 g/mol of O = 164 g/mol

Al₂(SO₄)₃ = 2*27 g/mol of Al + 3*32 g/mol of S + 12*16 g/mol of O = 342 g/mol

By the reaction stoichiometry:

3 moles of Ca(NO₃)₂ ---------------- 1 mol of Al₂(SO₄)₃

The mass is the number of moles multiplied by the molar mass, so:

3*164 of Ca(NO₃)₂ ------------------------ 1*342 of Al₂(SO₄)₃

492 g of Ca(NO₃)₂ --------------------- 342 of Al₂(SO₄)₃

90 g of Ca(NO₃)₂ --------------------- x

By a simple direct three rule:

492x = 30780

x = 62.56 g