Question

How many grams of CaSO4 (molar mass= 136.2g/m) can be produced from the reaction if 42.528 grams of CaBr2 is reacted with 42.39 grams of LiSO4?

Answer 1

The reaction of the calcium sulfate production is following:

$CaBr_2 + Li _2SO_4 = CaSO_4 + 2LiBr$

We need to determine the limiting reagent, which can be recognised by the lower mole number.

$n(CaBr_2)=\frac{m(CaBr_2)}{m(CaBr_2)}=\frac{42.528g}{199.89 g/mol} = 0.213 mol$

$n(Li_2SO_4) = \frac{42.39g}{109.94 g/mol}= 0.386 mol$

The CaBr₂ is limiting reagent and will participate in a reaction without excess.

Thus, the mass of CaSO₄ can be determined based on the equalities of the mole number:

$n(CaBr_2) = n(CaSO_4) = 0.213 mol$

$m(CaSO_4) = n\times M(CaSO_4) = 0.213 mol \times 136.2 g/mol = 29.01 g$