Question

19.3 g of cadmium hydroxide reacted with 15.21 g of hydrobromic acid. How many grams of water can be made?

Answer 1

`m_(H_2O)=3.384gH_2O`

Step-by-step explanation:

Hello,

In this case, the chemical reaction is:

`Cd(OH)_2+2HBr\rightarrow CdBr_2+2H_2O`

Thus, we first identify the limiting reactant by computing the yielded moles of water by both of the reactants:

`n_(H_2O)^(by\ Cd(OH)_2)=19.3gCd(OH)_2*(1molCd(OH)_2)/(146.4gCd(OH)_2)*(2molH_2O)/(1molCd(OH)_2)=0.264molH_2O\\\\n_(H_2O)^(by\ HBr)=15.21gHBr*(1molHBr)/(80.9gHBr)*(2molH_2O)/(2molHBr)=0.188molH_2O`

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:

`m_(H_2O)=0.188molH_2O*(18gH_2O)/(1molH_2O)\\ \\m_(H_2O)=3.384gH_2O`

Regards.