Na2CO3(aq) + CaCl2(aq) —> 2 NaCl(aq) + CaCO3(s) Calculate the volume (in mL) of 0.100 M Na2CO3 needed to produce 1.00 g of CaCO3(s). There is an excess of CaCl2.
Molar mass of calcium carbonate = 100.09 g/mol

100mL

Explanation:

To get the volume of CaCl2 we need to find the number of moles it will have firstly.

So using a mole-mass relationship between CaCl2 and CaCO3 we have;

1mole of CaCl2 produces 100.09g of CaCO3

The the amount of CaCl2 that'll produce 1g of CaCO3= 1×1/100.09=0.01mole.

To get our volume, we'll use the formula; n=CV where; n=number of mole,C=concentration and V=volume in dm^3 or liter.

Making volume subject of the formula we have; V=n/C

V=0.01/0.1=0.1L.

But since the answer is required to be in mL

So we have 0.1×1000=100mL.

I hope this helps :3

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