Answer:
[tex]m_{H_2O}=3.384gH_2O[/tex]
Explanation:
Hello,
In this case, the chemical reaction is:
[tex]Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O[/tex]
Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:
[tex]n_{H_2O}^{by\ Cd(OH)_2}=19.3gCd(OH)_2*\frac{1molCd(OH)_2}{146.4gCd(OH)_2}*\frac{2molH_2O}{1molCd(OH)_2}=0.264molH_2O\\\\n_{H_2O}^{by\ HBr}=15.21gHBr*\frac{1molHBr}{80.9gHBr}*\frac{2molH_2O}{2molHBr}=0.188molH_2O[/tex]
In such a way, since HBr yields less water than cadmium hydroxide, we infer that HBr is the limiting one, therefore, the yielded mass of water are:
[tex]m_{H_2O}=0.188molH_2O*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O[/tex]
Regards.
