Em tham khảo nha :
\(\begin{array}{l}
15)A\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{m_C} = \dfrac{{1000 \times 90}}{{100}} = 900g\\
{n_C} = \dfrac{{900}}{{12}} = 75mol\\
\Rightarrow {n_{C{O_2}}} = {n_C} = 75mol\\
{V_{C{O_2}}} = 75 \times 22,4 = 1680l\\
16)D\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{m_C} = \dfrac{{200 \times 69,5}}{{100}} = 139g\\
{n_C} = \dfrac{{139}}{{12}} = \dfrac{{139}}{{12}}mol\\
\Rightarrow {n_{C{O_2}}} = {n_C} = \dfrac{{139}}{{12}}mol\\
{m_{C{O_2}}} = \dfrac{{139}}{{12}} \times 44 = 509,67g\\
17)A\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{n_C} = \dfrac{6}{{12}} = 0,5mol\\
\Rightarrow {n_{C{O_2}}} = {n_C} = 0,5mol\\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,5mol\\
{m_{BaC{O_3}}} = 0,5 \times 100 = 50g\\
18)B\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
C + {O_2} \xrightarrow{t^0} C{O_2}\\
{n_{{O_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
hh:C(a\,mol);S(b\,mol)\\
a + b = 0,15(1)\\
12a + 32b = 2,8(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,05\\
{m_C} = 0,1 \times 12 = 1,2g\\
{m_S} = 2,8 - 1,2 = 1,6g
\end{array}\)
