Answer:
[tex]m_{CaCl_2}=39.96gCaCl_2[/tex]
Explanation:
Hello,
In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:
[tex]m_{CaCl_2}=14.4gCa*\frac{1molCa}{40gCa} *\frac{1molCaCl_2}{1molCa} *\frac{111gCaCl_2}{1molCaCl_2}\\ \\m_{CaCl_2}=39.96gCaCl_2[/tex]
Clearly, chlorine is not used since it is said there is enough for the reaction to go to completion.
Best regards.