\(n_{Al}^{} = \frac{{3,5}}{{27}} = \frac{7}{{54}}\;{\text{mol}}\)
\({\text{m}}_{H_2^{}SO_4^{}}^{} = 180.12,25\% = 22,05\;{\text{mol}}\)
\({\text{n}}_{H_2^{}SO_4^{}}^{} = 0,225\,\;{\text{mol}}\)
\(2Al + 3H_2^{}SO_4^{} \to Al_2^{}(SO_4^{})_3^{} + 3H_2^{}\)
\(\frac{7}{{54}}/2\;{\text{ < }}\frac{{0,225}}{3} \to \) \(H_2^{}SO_4^{}\) phản ứng hết, tính theo Al
\(\frac{7}{{54}}\;{\text{ }}\frac{7}{{36}}\,{\text{ }}\frac{7}{{108}}{\text{ }}\frac{7}{{36}}\)
\(V_{H_2^{}}^{} = \frac{7}{{36}}.22,4 = 4,35l\)
\(b)\)
\(BTKL:m_{Al}^{} + m_{{\text{dd}}H_2^{}SO_4^{}}^{} = m_{{\text{dd}}_{sauPU}^{}}^{} + m_{H_2^{}}^{}\)
\(3,5 + 180 = m_{{\text{dd}}_{SauPU}^{}}^{} + \frac{7}{{36}}.2\)
\( \to m_{{\text{dd}}_{sauPU}^{}}^{} = 183,1g\)
\(C\% _{Al_2^{}(SO_4^{})_3^{}}^{} = \frac{{\frac{7}{{108}}.342}}{{183,1}}.100 = 12,1\% \)
\(C\% _{H_2^{}SO_{4_{du}^{}}^{}}^{} = \frac{{(0,225 - \frac{7}{{36}}).98}}{{183,1}}.100 = 1,635\% \)
