For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of

Question

4 years ago

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For the following reaction, 5.22 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 12.9 grams of

aluminum sulfate. aluminum oxide (s) + sulfuric acid (aq) aluminum sulfate (aq) + water (l) What is the theoretical yield of aluminum sulfate ?

Chemistry

1 answer:

Mamont248 [21] · Answer 1 · 2022-01-17T13:26:52+03:00

Mamont248 [21]4 years ago

6 0

Answer:

$m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3$

Explanation:

Hello,

In this case, the undergoing balanced chemical reaction is:

$Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)$

Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:

$m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3$

Moreover, the percent yield is:

$Y=\frac{12.9g}{17.5g} *100\%=73.7\%$

Best regards.