For the reaction, calculate how many grams of the product form when 18.4g of Ca completely reacts.
Assume that there is more than enough of the other reactant.
The mol of Ca are:
then we get 0.46 mol of CaCl2, since the molecular mass of CaCl2 is 111, then we get:
0.46*111 = 51.06 grams of CaCl2
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