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Đáp án:

\(\begin{array}{l}
15,A\\
16,C\\
17,A\\
18,B\\
19,C\\
20,C\\
21,C\\
22,B
\end{array}\)

Giải thích các bước giải:

15,

\(\begin{array}{l}
2Cu + {O_2} \to 2CuO\\
{n_{Cu}} = 0,1mol\\
 \to {n_{CuO}} = {n_{Cu}} = 0,1mol\\
 \to m = {m_{CuO}} = 8g
\end{array}\)

16, 

\(\begin{array}{l}
R + {H_2}S{O_4} \to RS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,3mol\\
 \to {n_R} = {n_{{H_2}}} = 0,3mol\\
 \to {M_R} = 40\\
 \to Ca
\end{array}\)

17,

\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,5mol\\
 \to {n_{Fe}} = {n_{{H_2}}} = 0,5mol\\
 \to {m_{Fe}} = 28g
\end{array}\)

18,

\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = 0,1mol\\
 \to {n_{NaOH}} = {n_{Na}} = 0,1mol\\
 \to {m_{NaOH}} = 4g\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} = 0,05mol\\
 \to {m_{dd}} = 2,3 + 97,8 - 0,05 \times 2 = 100g\\
 \to C{\% _{NaOH}} = \dfrac{4}{{100}} \times 100\%  = 4\% 
\end{array}\)

19,

\(\begin{array}{l}
2M + C{l_2} \to 2MCl\\
{n_M} = {n_{MCl}}\\
 \to \dfrac{{4,6}}{M} = \dfrac{{11,7}}{{M + 35,5}}\\
 \to M = 23\\
 \to Na
\end{array}\)

20,

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
{n_{{H_2}}} = 0,45mol\\
 \to 1,5{n_{Al}} + {n_{Mg}} = {n_{{H_2}}} = 0,45mol\\
27{n_{Al}} + 24{n_{Mg}} = 9\\
\left\{ \begin{array}{l}
27{n_{Al}} + 24{n_{Mg}} = 9\\
1,5{n_{Al}} + {n_{Mg}} = 0,45
\end{array} \right.\\
 \to {n_{Al}} = 0,2mol \to {n_{Mg}} = 0,15mol\\
 \to \% {m_{Al}} = \dfrac{{0,2 \times 27}}{9} \times 100\%  = 60\% \\
 \to \% {m_{Mg}} = 100\%  - 60\%  = 40\% 
\end{array}\)

21,

\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{Fe}} = 0,4mol\\
 \to {n_{{H_2}S{O_4}}} = {n_{Fe}} = 0,4mol\\
 \to {m_{{H_2}S{O_4}}} = 39,2g\\
 \to C{\% _{{H_2}S{O_4}}} = \dfrac{{39,2}}{{200}} \times 100\%  = 19,6\% 
\end{array}\)

22,

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,4mol\\
 \to {n_{MgC{l_2}}} = {n_{Mg}} = 0,4mol\\
 \to {m_{MgC{l_2}}} = 38g\\
{n_{{H_2}}} = {n_{Mg}} = 0,4mol\\
 \to {m_{dd}} = 9,6 + 120 - 0,4 \times 2 = 128,8g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{38}}{{128,8}} \times 100\%  = 29,5\% 
\end{array}\)