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6.42g of Na2CO3 are mixed w/ 50mL of 3.0 M CaCl2 solution. What mass of Na2CO3 should be dissolved in the resulting solution?
The chemical equation for it is: Na2CO3 + CaCl2 = CaCO3 + 2 NaCl
2 Answers
- Trevor HLv 72 years agoFavorite Answer
From the balanced equation
Na2CO3 + CaCl2 = CaCO3 + 2 NaCl
1mol Na2CO3 reacts with 1 mol CaCl2
Molar mass Na2CO2 = 105.988 g/mol
mol Na2CO3 in 6.42g = 6.42g/105.988g/mol = 0.0605 mol
Mol CaCl2 in 50mL of 3.0M solution = 50mL / 1000mL/L * 3.0mol/L = 0.15 mol CaCl2
The Na2CO3 is the limiting reactant
There will be no Na2CO3 in the resulting solution.
- pisgahchemistLv 72 years ago
What mass.....
Na2CO3(s) + CaCl2(aq) --> CaCO3(s) + NaCl(aq)
6.42g ............ 50.0 mL
...................... 3.0M
0.050L x (3.0 mol CaCl2 / 1L) x (1 mol Na2CO3 / 1 mol CaCl2) x (106.0g Na2CO3 / 1 mol Na2CO3) = 15.9g Na2CO3
If all of the CaCl2 were to react, it would require 15.9g of Na2CO3. Since there is only 6.42 grams, the Na2CO3 is the limiting reactant, and all of the carbonate ends up as CaCO3. Only the Na+ ions are left over.
