Bài 1:

a)m$Fe_2$$O_3$=n.M=0,15.160=24 (g)

b)mMgO=n.M=0,75.40=30 (g)

Bài 2:

a)nCuO=$\frac{m}{M}$=$\frac{2}{80}$=0,025 (mol)

b)nNaOH=$\frac{m}{M}$=$\frac{10}{40}$=0,25 (mol)  

Bài 3:

a)V$O_2$(đktc)=n.22,4=0,5.22,4=11,2 (l)

b)V$CO_2$(đktc)=n.22,4=1,25.22,4=28 (l)

Bài 4:

a)n$H_2$(đktc)=$\frac{V}{22,4}$=$\frac{4,48}{22,4}$=0,2 (mol)

b)n$CH_4$(đktc)=$\frac{V}{22,4}$=$\frac{67,2}{22,4}$=3 (mol)

c)nFe=$\frac{m}{M}$=$\frac{28}{56}$=0,5 (mol)

d)nAl=$\frac{m}{M}$=$\frac{5,4}{27}$=0,2 (mol)    

Bài 5:

$M_A$=$\frac{m}{n}$=$\frac{15,5}{0,25}$=62

⇔2R+O=62

⇔2R+16=62

⇔2R=46

⇔R=23

Mà kim loại có nguyên tử bằng 23 là Natri (Na)

Vậy A có CTHH là $Na_2O$

Em tham khảo nha :

\(\begin{array}{l}
1)\\
{m_{F{e_2}{O_3}}} = 0,15 \times 160 = 24g\\
{m_{MgO}} = 0,75 \times 40 = 30g\\
2)\\
{n_{CuO}} = \dfrac{2}{{80}} = 0,025mol\\
{n_{NaOH}} = \dfrac{{10}}{{40}} = 0,25mol\\
3)\\
{V_{{O_2}}} = 0,5 \times 22,4 = 11,2l\\
{V_{C{O_2}}} = 1,25 \times 22,4 = 28l\\
4)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{C{H_4}}} = \dfrac{{67,2}}{{22,4}} = 3mol\\
{n_{Fe}} = \dfrac{{28}}{{56}} = 0,5mol\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
5)\\
CTHH:{R_2}O\\
{M_{{R_2}O}} = \dfrac{{15,5}}{{0,25}} = 62dvC\\
2{M_R} + {M_O} = 62\\
 \Rightarrow {M_R} = \dfrac{{62 - 16}}{2} = 23dvC\\
R:Natri(Na)\\
CTHH:N{a_2}O
\end{array}\)