Đáp án:

 33,33% và 66,67%

8g

Giải thích các bước giải:

\(\begin{array}{l}
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
2{C_3}{H_6} + 9{O_2} \to 6C{O_2} + 6{H_2}O\\
nhh = \frac{{0,672}}{{22,4}} = 0,03\,mol\\
n{O_2} = \frac{{2,688}}{{22,4}} = 0,12\,mol\\
hh:{C_2}{H_4}(a\,mol),{C_3}{H_6}(b\,mol)\\
a + b = 0,03\\
3a + 4,5b = 0,12\\
 =  > a = 0,01\,b = 0,02\\
\% V{C_2}{H_4} = 33,33\% \\
\% V{C_3}{H_6} = 100 - 33,33 = 66,67\% \\
nC{O_2} = 0,01 \times 2 + 0,02 \times 3 = 0,08\,mol\\
nCaC{O_3} = nC{O_2} = 0,08\,mol\\
mCaC{O_3} = 0,08 \times 100 = 8g
\end{array}\)

$n_{Khí}=\dfrac{0,672}{22,4}=0,03(mol)\\ n_{O_2}=\dfrac{2,688}{22,4}=0,12(mol)\\ Đặt\ \begin{cases}n_{C_2H_4}=x(mol)\\n_{C_3H_6}=y(mol)\end{cases}\\ \Rightarrow x+y=003\ (1)\\ C_2H_4+3O_2\xrightarrow{t^0} 2CO_2+2H_2O\\ C_3H_6+\dfrac{9}{2}O_2\xrightarrow{t^0} 3CO_2+3H_2O\\ \xrightarrow{\text{Từ PTHH}}\ n_{O_2}=2.n_{C_2H_4}+\dfrac{9}{2}.n_{C_3H_6}=3x+\dfrac{9}{2}y(mol)\\ \Rightarrow 2x+\dfrac{9}{2}y=0,12\ (2)\\ Từ \ (1);(2)\ \Rightarrow \begin{cases}x=0,01\\y=0,02\end{cases}\\ \%V_{C_2H_4}=\dfrac{0,01}{0,03}.100\%=33,33\%\\ \%V_{C_3H_6}=\dfrac{0,02}{0,03}.100\%=66,67\%\\ \xrightarrow{\text{BTNT C}}\ n_{CaCO_3}=2.n_{C_2H_4}+3.n_{C_3H_6}=0,08(mol)\\ m=0,08.100=8$