Đáp án:

\(\begin{array}{l} b,\ \%m_{Al}=20,69\%\\ \%m_{K_2CO_3}=79,31\%\\ c,\ m_{\text{dd H$_2$SO$_4$}}=45\ g.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow\ (1)\\ K_2CO_3+H_2SO_4\to K_2SO_4+CO_2\uparrow+H_2O\ (2)\\ CO_2+Ca(OH)_2\to CaCO_3\downarrow+H_2O\ (3)\\ b,\\ n_{H_2}=\dfrac{1,344}{22,4}=0,06\ mol.\\ n_{CaCO_3}=\dfrac{3}{100}=0,03\ mol.\\ Theo\ pt\ (3):\ n_{CO_2}=n_{CaCO_3}=0,03\ mol.\\ Theo\ pt\ (2):\ n_{K_2CO_3}=n_{CO_2}=0,03\ mol.\\ Theo\ pt\ (1):\ n_{Al}=\dfrac{2}{3}n_{H_2}=0,04\ mol.\\ \Rightarrow \%m_{Al}=\dfrac{0,04\times 27}{0,04\times 27+0,03\times 138}\times 100\%=20,69\%\\ \%m_{K_2CO_3}=\dfrac{0,03\times 138}{0,04\times 27+0,03\times 138}\times 100\%=79,31\%\\ c,\\ \sum n_{H_2SO_4}=n_{H_2}+n_{CO_2}=0,06+0,03=0,09\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,09\times 98}{19,6\%}=45\ g.\end{array}\)

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