Em tham khảo nha :

\(\begin{array}{l}
a)\\
{m_{N{a_2}S{O_3}}} = 0,3 \times 126 = 37,8g\\
{n_{C{O_2}}} = \dfrac{{247,9}}{{24,79}} = 10mol\\
{m_{C{O_2}}} = 10 \times 44 = 440g\\
{m_{CaC{l_2}}} = 0,05 \times 111 = 5,55g\\
b)\\
{n_{NaCl}} = \dfrac{{1,8 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,3mol\\
{n_{NO}} = \dfrac{{7,437}}{{24,79}} = 0,3mol\\
{n_{Zn}} = \dfrac{{13}}{{65}} = 0,2mol\\
c)\\
{n_{C{O_2}}} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{V_{C{O_2}}} = 0,2 \times 24,79 = 4,958l\\
{n_{{O_2}}} = \dfrac{{7,2}}{{32}} = 0,225mol\\
{V_{{O_2}}} = 0,225 \times 24,79 = 5,57775l\\
{n_{{O_2}}} = \dfrac{{0,6}}{2} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 24,79 = 7,437l
\end{array}\)

 

Đáp án+Giải thích các bước giải:

 a.

$-$ $m_{Na_2SO_3}$ = 0,3 × 126 = 37,8 (g)

 $m_{CO_2(đkc)}$ = $\frac{247,9}{24,79}$ = 10 (mol)

→ $m_{CO_2}$ = 10 × 44 = 440 (g)

$-$  $m_{CaCl_2}$ = 0,05 × 75,5 = 3,755 (g)

b. 

$-$ $n_{NaCl}$ = $\frac{1,8×10^{23}}{6×10^{23}}$ = 0,3 (mol)

$-$ $n_{NO}$ = $\frac{7,437}{24,79}$ = 0,3 (mol)

$-$ $n_{Zn}$ = $\frac{13}{65}$ = 0,2 (mol)

c. 

$-$ $n_{CO_2}$ = $\frac{8,8}{44}$ = 0,2 (mol)

→ $V_{CO_2(đkc)}$ = 0,2 × 24,79 = 4,958 (l)

$-$ $n_{O_2}$ = $\frac{7,2}{32}$ = 0,225 (mol)

→ $V_{O_2(đkc)}$ = 0,225 × 24,79 = 5,7775 (l)

$-$ $n_{H_2}$ = $\frac{0,6}{2}$ = 0,3 (mol)

→ $V_{O_2(đkc)}$ = 0,3 × 24,79 = 7,437 (l)