Answer:
NaCl is the limiting reactant
18.0g of CaCO3 are in excess
113.4g Na2CO3 are produced
118.7g CaCl2 are produced
Explanation:
To solve this question we need to convert the mass of each reactant to moles in order to find the limitng and excess reactant to find the amount of products that would be produced:
<em>Moles CaCO3:</em>
125g * (1mol / 100.09g) = 1.25 moles
<em>Moles NaCl:</em>
125g * (1mol / 58.44g) = 2.14 moles
For a complete reaction of 2.14 moles of NaCl are required:
2.14 moles NaCl * (1mol CaCO3 / 2 mol NaCl) = 1.07 moles of CaCO3.
As there are 1.25 moles, <em>Calcium carbonate is the excess reactant and NaCl the limiting reactant</em>
The moles of CaCO3 in excess are:
1.25mol - 1.07mol = 0.18mol CaCO3 and the mass is:
0.18mol CaCO3 * (100.09g / mol) = 18.0g of CaCO3 are in excess
The moles of Na2CO3 and CaCl2 produced are:
2.14 moles NaCl * (1mol Na2CO3-1mol CaCl2 / 2 mol NaCl) = 1.07 moles are produced.
The masses are:
1.07 mol Na2CO3 * (105.99g / mol) = 113.4g Na2CO3
1.07 mol CaCl2 * (110.98g / mol) = 118.7g CaCl2
