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$"Moles of calcium"$ $=$ $(1200xx10^-3*g)/(40.08*g*mol^-1)$ $=$ $0.04008*mol$.

And thus we need $0.04008*mol$ $"calcium carbonate"$, i.e. a mass of $0.04008*molxx100.09*g*mol^-1$ $=$ $??g$.

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