Answer:
15.2 grams of calcium chloride are produced and HCl is the limiting reactant.
Explanation:
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In this case, according to the described scenario, it is possible to realize that the reaction between hydrochloric acid and calcium hydroxide is:
[tex]2HCl+Ca(OH)_2\rightarrow CaCl_2+2H_2O[/tex]
Whereas there is a 2:1 mole ratio of the acid to the base. In such a way, with the given masses, we can compute how much calcium chloride product is produced due to the chemical reaction via stoichiometry:
[tex]m_{CaCl_2}^{by HCl}=10.0gHCl*\frac{1molHCl}{36.46gHCl}*\frac{1molCaCl_2}{2molHCl} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.2gCaCl_2\\\\m_{CaCl_2}^{by Ca(OH)_2}=10.5gHCl*\frac{1molCa(OH)_2}{74.09gCa(OH)_2}*\frac{1molCaCl_2}{1molCa(OH)_2} *\frac{110.98gCaCl_2}{1molCaCl_2} =15.7gCaCl_2[/tex]
Whereas we infer that the correct amount is 15.2 g since HCl is the limiting reactant as it produces the fewest grams of the desired product. Consequently, the calcium hydroxide is the excess reactant here.
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