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Musya8 [376]
3 years ago
7

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

Chemistry
1 answer:
Alecsey [184]3 years ago
4 0

Answer:

m_{H_2O}=3.384gH_2O

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*\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

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*\frac{18gH_2O}{1molH_2O}\\ \\m_{H_2O}=3.384gH_2O

Regards.

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