Câu 1

a.

`n_{H_2O}`=$\dfrac{6×10^23}{6×10^23}$ =`1 mol`

b.

`n_{Fe}`=$\dfrac{9×10^23}{6×10^23}$=`1,5 mol`

c.

`n_{O_2}`=$\dfrac{6,1975}{24,79}$=`0,25 mol`

d.

Đổi : `7437 ml= 7,437 l`

`n_{N_2}`=$\dfrac{7,437}{24,79}$=`0,3 mol`

Câu 2

a.

`V_{SO_2}`=`0,5×24,79=12,395 l`

b.

`V_{CO_2}`=`0,1×24,79=2,479 l`

c.

`n_{H_2}`=$\dfrac{6×10^23}{6×10^23}$=`1 mol`

`V_{H_2}`=`1×24,79=24,79 l`

d.

`n_{NO}`=$\dfrac{3×10^23}{6×10^23}$=`0,5 mol`

`V_{NO}`=`0,5×24,79=12,395 l`

Câu 3

a.Số nguyên tử `Al` = `0,5×6×10^23`=`3×10^23`

b.Số phân tử `NaCl` = `2×6×10^23`=`12×10^23`

c.

`n_{H_2}`=$\dfrac{24,79}{24,79}$=`1 mol`

Số phân tử `H_2` = `1×6×10^23=6×10^23`

d.

`n_{O_2}`=$\dfrac{49,58}{24,79}$=`2 mol`

Số phân tử `O_2` = `2×6×10^23`=`12×10^23`

Em tham khảo nha :

\(\begin{array}{l}
1)\\
{n_{{H_2}O}} = \dfrac{{6 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1mol\\
{n_{Fe}} = \dfrac{{9 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1,5mol\\
{n_{{O_2}}} = \dfrac{{6,1975}}{{24,79}} = 0,25mol\\
{n_{{N_2}}} = \dfrac{{7,437}}{{24,79}} = 0,3mol\\
2)\\
{V_{S{O_2}}} = 0,5 \times 24,79 = 12,395l\\
{V_{C{O_2}}} = 0,1 \times 24,79 = 2,479l\\
{n_{{H_2}}} = \dfrac{{6 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1mol\\
{V_{{H_2}}} = 1 \times 24,79 = 24,79l\\
{n_{NO}} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5mol\\
{V_{NO}} = 0,5 \times 24,79 = 12,395l\\
3)\\
{N_{Al}} = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}}\text{(Nguyên tử)}\\
{N_{NaCl}} = 2 \times 6 \times {10^{23}} = 12 \times {10^{23}}\text{(Nguyên tử)}\\
{n_{{H_2}}} = \dfrac{{24,79}}{{24,79}} = 1mol\\
{N_{{H_2}}} = 1 \times 6 \times {10^{23}} = 6 \times {10^{23}}\text{(Phân tử)}\\
{n_{{O_2}}} = \dfrac{{49,58}}{{24,79}} = 2mol\\
{N_{{O_2}}} = 2 \times 6 \times {10^{23}} = 12 \times {10^{23}}\text{(Phân tử)}
\end{array}\)