Giải thích các bước giải:

 Bài 1:
1, Carbon + Oxygen ---> Carbon dioxide
2, Aluminium + Sulfuric acid ---> Aluminium sulfate + Hydrogen
3, Iron(III) hydroxide + Hydrochloric acid ---> Iron(III) chloride + Nước
4, Phophorus + Oxygen ---> Phosphorus (V) oxide
5, Ethylen + Oxygen ---> Carbon dioxide + Hơi nước
Bài 2:
1, $n_{K2SO4}$ = $\frac{150.0,232}{39.2 + 32 + 16.4}$ = 0.2 (mol)
2,$n_{BaCl2}$ = $\frac{m}{M}$ = $\frac{120.0,26}{137+35,5.2}$ =0.15(mol)
3, $n_{NaOH}$ = $\frac{m}{M}$ = $\frac{50.0,2}{23+16+1}$ =0.25(mol)
4, $n_{NaNO3}$ = $\frac{m}{M}$ = $\frac{150.0,17}{23+14+16.3}$ = 0.3(mol)
Bài 3:
1, Ta có: 200ml = 0.2l
$n_{CuSO4}$ = $\frac{m}{M}$ = $\frac{32}{64+32+16.4}$ =0.2(mol)

=> $C_{M}$ = $\frac{0.2}{0.2}$ = 1M
2, Ta có: 150ml = 0.15l
$n_{MgCl2}$ = $\frac{m}{M}$ = $\frac{28,5}{24+35,5.2}$ =0,3(mol)
=>$C_{M}$ = $\frac{0.3}{0.15}$ = 2M
3, Ta có: 250ml=0.25l
$n_{Na2CO3}$ = $\frac{m}{M}$ = $\frac{53}{23.2+12+16.3}$ =0.5(mol)
=>$C_{M}$ = $\frac{0.5}{0.25}$ = 2M
4, Ta có: 500ml = 0.5l
$n_{Ca(OH)2}$ = $\frac{m}{M}$ = $\frac{148}{40+(16.1).2}$ =2(mol)
=>=>$C_{M}$ = $\frac{2}{0.5}$ = 4M

Đáp án + Giải thích các bước giải:

Bài 1:

1. Carbon + Không khí $\to$ Carbon dioxide.

2. Aluminium + Sulfuric acid $\to$ Aluminium sulfate + Hydrogen.

3. Iron(III) hydroxide + Hydrochloric acid $\to$ Iron(III) chloride + Nước

4. Phosphorus + Oxygen $\to$ Phosphorus(V) oxide.

5. Etilen + Không khí $\to$ Carbon dioxide + Hơi nước.

Bài 2:

Công thức $n=\dfrac{m_{dd}.C}{M}$

1. $n_{K_2SO_4}=\dfrac{150.23,2\%}{174}=0,2(mol)$

2. $n_{BaCl_2}=\dfrac{120.26\%}{208}=0,15(mol)$

3. $n_{NaOH}=\dfrac{50.20\%}{40}=0,25(mol)$

4. $n_{NaNO_3}=\dfrac{150.17\%}{85}=0,3(mol)$

Bài 3:

Công thức: $C_{M}=\dfrac{n}{V_{dd}}$

1. $n_{CuSO_4}=\dfrac{32}{160}=0,2(mol)$

$ \to C^{M}_{CuSO_4}=\dfrac{0,2}{200.10^{-3}}=1(M)$

2. $n_{MgCl_2}=\dfrac{28,5}{95}=0,3(mol)$

$ \to C^{M}_{MgCl_2}=\dfrac{0,3}{150.10^{-3}}=2(M)$

3. $n_{Na_2CO_3}=\dfrac{53}{106}=0,5(mol)$

$ \to C^{M}_{Na_2CO_3}=\dfrac{0,5}{250.10^{-3}}=2(M)$

4. $n_{Ca(OH)_2}=\dfrac{148}{74}=2(mol)$

$ \to C^{M}_{Ca(OH)_2}=\dfrac{2}{5.10^{-3}}=4(M)$