Question

Calculate the grams of solid product and excess reagent when 2.50 grams of barium chloride is mixed with 4.00 grams of magnesium sulfate.

Answer 1

Given

Mass of BaCl2 = 2.5 g

Mass of MgSO4 = 4.0 g

The reaction will be

BaCl2(aq) + MgSO4(aq) → BaSO4(s) + MgCl2 (aq)

The solid formed will be BaSO4

Lets calculate no of moles of BaCl2

No of moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2

= 2.5 g / 208.23 g/mol = 0.012 moles BaCl2

Now

No of moles of MgSO4 = mass of MgSO4 / molar mass of MgSO4

= 4 g / 120.37 g/mol = 0.033 moles MgSO4

Now as the no of moles of BaCl2 is less than MgSO4 it will be the limiting reagent

Total moles (0.012) of BaCl2 will be reacted with 0.012 moles of MgSO4 to form BaSO4

So

From reaction we can say that

1 mole of BaCl2 produces 1 mole of Solid BaSo4

Hence

0.012 moles BaCl2 produces 0.012 moles BaSO4

Mass of BaSO4 = moles of BaSO4 * molar Mass of BaSO4

= 0.012 mol * 233.38 g/mol = 2.8 g of BaSO4

Also excess reagent is MgSO4

From the reaction we can say that

1 mole of BaCl2 requires 1 mole of MgSO4

Hence

0.012 moles BaCl2 requires 0.012 moles MgSO4

Mass of MgSO4 reacted = no of moles * molar mass

= 0.012 moles * 120.37 g/mol = 1.5 g of MgSO4 is reacted

So the remaining moles of MgSO4 is

0.033 moles - 0.012 moles = 0.021 moles

Mass of MgSO4 remaining

= no of moles of MgSO4 * molar mass of MgSO4

= 0.021 moles * 120.37 g/mol = 2.5 g of MgSO4

Hence

Mass of solid product i.e BaSO4 is 2.8 g

Mass of remaining excess reagent i.e MgSO4 is 2.5 g