Đáp án:
a) 2,4g và 10g
b) 0,793
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
nhhY = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh:Mg(a\,mol),CaC{O_3}(b\,mol)\\
a + b = 0,2\\
24a + 100b = 12,4\\
\Rightarrow a = b = 0,1\,mol\\
\% mMg = \dfrac{{0,1 \times 24}}{{12,4}} \times 100\% = 19,35\% \\
\% mCaC{O_3} = 100 - 19,35 = 80,65\% \\
b)\\
n{H_2} = nMg = 0,1\,mol\\
nC{O_2} = nCaC{O_3} = 0,1\,mol\\
Mhh = \dfrac{{0,1 \times 2 + 0,1 \times 44}}{{0,2}} = 23g/mol\\
\Rightarrow dY/kk = \dfrac{{23}}{{29}} = 0,793\\
c)\\
nHCl = \dfrac{{100 \times 18,25\% }}{{36,5}} = 0,5\,mol\\
nHCl\,spu = 0,5 - 0,2 \times 2 = 0,1\,mol\\
m{\rm{dd}}spu = 12,4 + 100 - 0,1 \times 2 - 0,1 \times 44 = 107,8g\\
C\% MgC{l_2} = \dfrac{{0,1 \times 95}}{{107,8}} \times 100\% = 8,81\% \\
C\% CaC{l_2} = \dfrac{{0,1 \times 111}}{{107,8}} \times 100\% = 10,3\% \\
C\% HCl = \dfrac{{0,1 \times 36,5}}{{107,8}} \times 100\% = 3,39\%
\end{array}\)
