2)

Ta có:

a)\({m_{CuS{O_4}}} = 0,15.(64 + 32 + 16.4) = 24{\text{ gam}}\)

b)

\({n_{C{O_2}}} = \frac{{5,6}}{{22,4}} = 0,25{\text{ mol}}\)

\( \to {m_{C{O_2}}} = 0,25.(12 + 16.2) = 11{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(2Fe + 3C{l_2}\xrightarrow{{{t^o}}}2FeC{l_3}\)

\({P_2}{O_5} + 3{H_2}O\xrightarrow{{}}2{H_3}P{O_4}\)

\(C{O_2} + 2KOH\xrightarrow{{}}{K_2}C{O_3} + {H_2}O\)

\(A{l_2}{O_3} + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}O\)

4)

Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Muối kẽm là \(ZnCl_2\).

BTKL:

\({m_{Zn}} + {m_{HCl}} = {m_{ZnC{l_2}}} + {m_{{H_2}}}\)

\( \to 6,5 + 7,3 = {m_{ZnC{l_2}}} + 0,2 \to {m_{ZnC{l_2}}} = 13,6{\text{ gam}}\)

Bài giải:

Câu 2:

-$m_{CuSO_4}=n_{CuSO_4}.M_{CuSO_4}$

          $=0,15.(64+32+16.4)=24(g)$

-$n_{CO_2}=\frac{V_{CO_2}}{22,4}=\frac{5,6}{22,4}=0,25(mol)$ 

$m_{CO_2}=n_{CO_2}.M_{CO_2}$

           $=0,25.(12+16.2)=11(g)$

Câu 3:

1. $2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3 $

2. $P_2O_5+3H_2O→2H_3PO_4$

3. $CO_2+2KOH→K_2CO_3+H_2O$

4. $Al_2O_3+6HCl→2AlCl_3+3H_2O$

Câu 4:

-Công thức muối kẽm clorua: $ZnCl_2$

-Áp dụng định luật bảo toàn khối lượng

$⇒m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}$

$⇔6,5+7,3=m_{ZnCl_2}+0,2$

$⇔m_{ZnCl_2}=6,5+7,3-0,2=13,6(g)$