Đáp án:

Giải thích các bước giải:

Câu 1: Chọn B.

$n_{CuO}$= $\frac{16}{80}$= $0,2(mol)$

$Pt: C + 2CuO$ $\xrightarrow{t^o}$ $2Cu + CO_2(1)$

$\text{Từ pt(1):}$ $n_{C}$= $0,1(mol)$

$\rightarrow$ $m_{C}$= $0,1.12= 1,2(g)$

Câu 2: Chọn D.

$6KOH + 3Cl_2$ $\rightarrow$ $5KCl + KClO_3 + 3H_2O(1)$

$n_{Cl_2}$= $\frac{2,24}{22,4}$= $0,1(mol)$

$\text{Từ pt(1):}$ $n_{KOH}$= $0,2(mol)$

$\rightarrow$ $V_{KOH}$= $\frac{0,2}{2}$= $0,1(l)$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
B\\
C + 2CuO \to 2Cu + C{O_2}\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_C} = \dfrac{{{n_{CuO}}}}{2} = 0,1mol\\
{m_C} = 0,1 \times 12 = 1,2g\\
2)\\
D\\
2KOH + C{l_2} \to KCl + KClO + {H_2}O\\
{n_{C{l_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{KOH}} = 2{n_{C{l_2}}} = 0,2mol\\
{V_{KOH}} = \dfrac{{0,2}}{2} = 0,1l
\end{array}\)