Answer to Question #135935 in General Chemistry for Daisha-Jade Gaspar

Question

If you have 3.16 moles of Na2CO3 and 1.99 moles of Ca(HC2H3O2)2, how many moles of NaHC2H3O2 will be produced?

Answer 1

The main reaction between calcium acetate and sodium carbonate is

Na2Co3 + Ca(CH3COO)2 --> CaCo3 + 2Na(CH3COO)

So here from reaction it is clear that 1mole of SodiumCarbonate react with one mole of CalciumAcetate and will form 2 Moles of Sodiumacetate ....

In our question Mole of Na2CO3 = 3.16

MOLES OF Calcium Acetate= 1.99

Hence limiting reagent is Calcium acetate

SO mole of Sodium acetate will be 2x1.99 = 3.98 Moles.

Answer 3.98 Moles Sodiumacetate will be formed