Question

Use the balanced equation to solve the problem.Na₂CO3 +CaCl2 → 2NaCl + CaCO3-122g Na₂CO3 and 95.0g CaCl2 are mixed in solution and react.How many moles Na₂CO3 are available to react?mol Na₂CO3How many moles CaCl₂ are available to react?mol CaCl₂What is the limiting reactant?How many grams of CaCO3 are made?CaCl₂g CaCO3

Answer 1

Explanation:

First, let's rewrite the balanced equation here:

Na₂CO3 +CaCl2 → 2NaCl + CaCO3

Now let's transform 122 g of Na2CO3 into moles and 95 g of CaCl2 into moles as well. For this, we use the following formula: n = m/MM

First, for Na2CO3:

MM of Na2CO3 = 106 g/mol

m of Na2CO3 = 122 g

n = 122/106

n = 1.15 moles

For CaCl2:

MM of CaCl2 = 111 g/mol

m of CaCl2 = 95

n = 95/111

n = 0.856 moles

CaCl2 is the limiting reactant, so 0.856 moles of it is going to react with 0.856 moles of Na2CO3 and produce 0.856 moles of CaCO3.

Let's transform 0.856 moles of CaCO3 into grams:

MM of CaCO3 = 100 g/mol

m = n*MM

m = 0.856*100

m = 85.6 g

Answer:

How many moles Na2CO3 are available to react?

1.15 mol of Na2CO3

How many moles CaCl₂ are available to react?

0.856 mol of CaCl2

What is the limiting reactant?

The limiting reactant is CaCl2.

How many grams of CaCO3 are made?

85.6 g