Đáp án + Giải thích các bước giải:

* $Na_{2}$$ZnO_{2}$

n$Na_{2}$$ZnO_{2}$= $\frac{m}{M}$ = $\frac{8.58}{143}$ = 0.06 (mol)

* $Mg_{3}$$(PO4)_{2}$ 

n $Mg_{3}$$(PO4)_{2}$= $\frac{m}{M}$=$\frac{7.86}{262}$ = 0.03 (mol)

* $CuSO_{4}$ 

n$CuSO_{4}$= $\frac{m}{M}$ = $\frac{7.42}{160}$ = 0.046375 (mol)
* $CO_{2}$

n $CO_{2}$ = $\frac{V}{22.4}$ = $\frac{26.88}{22.4}$ = 1.2 (mol)

* $NO_{}$

n $NO_{}$ = $\frac{V}{22.4}$=  $\frac{20.16}{22.4}$ = 0.9 (mol)

* $NO_{2}$ 

n $NO_{2}$ = $\frac{V}{22.4}$ $\frac{16.8}{22.4}$ = 0.75 (mol)

* $CuSO_{4}$ 

n $CuSO_{4}$ = $\frac{sptu}{6.10²³}$ = $\frac{1.2×10²³}{6.10²³}$ = 0.2 (mol)

* $Fe_{}$ 

n$Fe_{}$ = $\frac{sptu}{6.10²³}$ = $\frac{4.8×10²²}{6.10²³}$ = 0.08 (mol)

* $Na_{3}$$PO_{4}$

n $Na_{3}$$PO_{4}$ = $\frac{sptu}{6.10²³}$ = $\frac{15.3×10²³}{6.10²³}$ = 2.55 (mol)

#milkteanguyen

Em tham khảo nha :

\(\begin{array}{l}
{n_{N{a_2}Zn{O_2}}} = \dfrac{{8,58}}{{143}} = 0,06mol\\
{n_{CuS{O_4}}} = \dfrac{{7,42}}{{160}} = 0,046375mol\\
{n_{M{g_3}{{(P{O_4})}_2}}} = \dfrac{{7,86}}{{262}} = 0,03mol\\
{n_{C{O_2}}} = \dfrac{{26,88}}{{22,4}} = 1,2mol\\
{n_{NO}} = \dfrac{{20,16}}{{22,4}} = 0,9mol\\
{n_{N{O_2}}} = \dfrac{{16,8}}{{22,4}} = 0,75mol\\
{n_{CuS{O_4}}} = \dfrac{{1,2 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,2\\
{n_{Fe}} = \dfrac{{4,8 \times {{10}^{22}}}}{{6 \times {{10}^{23}}}}0,08mol\\
{n_{N{a_3}P{O_4}}} = \dfrac{{15,3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 2,55mol
\end{array}\)