Find how much of the solid product if formed?? Please help
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Find how much of the solid product if formed?? Please help

[From: ] [author: ] [Date: 13-11-05] [Hit: ]
.0426 mol CuCO3 X (123.56 g /mol) = 5.26 g of solid preciptate.......
10.0 grams of sodium carbonate is added to 50 mL of water. Then 8.00 grams of copper (I) nitrate are added to 50 mL of water. The two solutions are then poured together and stirred. Calculate how much of the solid product is formed.

I don't even know where to start please show me how you got this.

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The formula is : Cu(NO3)(aq) + Na2CO3(aq) ---------> CuCO3(s) + 2 NaNO3 (aq)

10.0 g Na2CO3 X (1 mol/105.99 g) = .0943 mol Na2CO3
8.00 Cu(NO3) X (1 mol/187.86 g) = .0426 mol Cu(NO3)

because the reaction is a one to one ratio of either reactant to product, that would make the reactant with the least number of mols the limiting reactant. So the limiting reactant is the Cu(NO3).

If you multiply with stoichiometry

.0426 mol Cu(NO3) X (1 mol CuCO3/1 mol Cu(NO3)) = .0426 mol CuCO3

making .0426 mol CuCO3 and if you need it in grams

.0426 mol CuCO3 X (123.56 g /mol) = 5.26 g of solid preciptate.
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