Calculate the volume (in mL) of 0.100 M Na, CO3 needed to produce 1.00 g of CaCO3(s)
. There is an excess of CaCl2.
Molar mass of calcium carbonate = 100.09 g/mol

10 ml

Explanation:

Molarity is equal to moles/volume (Liters) so you need to find how many moles are in 1.00 gram of Calcium carbonate

100.09 MM = 1.00/Xmol

move the equation so that mol is on the left

Xmol = 1.00 g/ 100.09 MM = .009991 mol (rounding = .01 mol)

then add that into your equation to find volume

M = molarity

mol= moles

V = volume in Liters

M= mol/V so 1.00M = .01 mol/ v

V = 0.01 L

since the answer is in Liters you need to multiply that answer by 1000 which is the conversion from Liters to mililiters

.01 L X1000 = 10 ml

it means there are 3 atoms of oxygen