Em tham khảo nha :

\(\begin{array}{l}
6)A\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to 2N{H_4}N{O_3} + A{g_2}{C_2}\\
{n_{{C_2}{H_2}}} = \dfrac{{5,2}}{{26}} = 0,2mol\\
{n_{A{g_2}{C_2}}} = {n_{{C_2}{H_2}}} = 0,2mol\\
{m_{A{g_2}{C_2}}} = 0,2 \times 240 = 48g\\
7)A\\
CH \equiv C - C{H_3} + AgN{O_3} + N{H_3} \to AgC \equiv C - C{H_3} + N{H_4}N{O_3}\\
{n_{{C_3}{H_3}Ag}} = \dfrac{{14,7}}{{147}} = 0,1mol\\
{n_{{C_3}{H_4}}} = {n_{{C_3}{H_3}Ag}} = 0,1mol\\
{m_{{C_3}{H_4}}} = 0,1 \times 40 = 4g
\end{array}\)

Câu 6:

$n_{C_2H_2}=\dfrac{5,2}{26}=0,2(mol)$

$C_2H_2\xrightarrow{{AgNO_3/NH_3}} Ag_2C_2$

$\to n_{Ag_2C_2}=n_{C_2H_2}=0,2(mol)$

$\to m=0,2.240=48g$

Chọn $A$

Câu 7:

$n_{C_3H_3Ag\downarrow }=\dfrac{14,7}{147}=0,1(mol)$

$CH\equiv C-CH_3\xrightarrow{{AgNO_3/NH_3}} AgC\equiv C-CH_3$

$\to n_{C_3H_4}=0,1(mol)$

$\to m=0,1.40=4g$

Chọn $A$