Giải thích các bước giải:

`a.`

`-n_{Fe_2(SO_4)_3}=\frac{m_{Fe_2(SO_4)_3}}{M_{Fe_2(SO_4)_3}}=\frac{60}{400}=0,15(mol)`

Bảo toàn nguyên tố `Fe:`

`-n_{Fe}=2.n_{Fe_2(SO_4)_3}=2.0,15=0,3(mol)`

`→m_{Fe}=n_{Fe}.M_{Fe}=0,3.56=16,8(g)`

Bảo toàn nguyên tố `S:`

`-n_{S}=3.n_{Fe_2(SO_4)_3}=3.0,15=0,45(mol)`

`→m_S=n_S.M_S=0,45.32=14,4(g)`

`→m_O=m_{Fe_2(SO_4)_3}-m_{Fe}-m_S`

`→m_O=60-16,8-14,4=28,8(g)`

`b.`

`-n_{N_2O}=\frac{V_{N_2O}}{22,4}=\frac{3,36}{22,4}=0,15(mol)`

Bảo toàn nguyên tố `N:`

`-n_N=2.n_{N_2O}=2.0,15=0,3(mol)`

`→m_N=n_N.M_N=0,3.14=4,2(g)`

Bảo toàn nguyên tố `O:`

`-n_O=n_{N_2O}=0,15(mol)`

`→m_O=n_O.M_O=0,15.16=8(g)`

`c.`

Số Avogađro `1(mol)≈6.10^{23}` phân tử 

`-n_{MgCO_3}=\frac{1,8.10^{22}}{6.10^{23}}=0,03(mol)`

Bảo toàn nguyên tố `Mg:`

`-n_{Mg}=n_{MgCO_3}=0,03(mol)`

`→m_{Mg}=n_{Mg}.M_{Mg}=0,03.24=0,72(g)`

Bảo toàn nguyên tố `C:`

`-n_C=n_{MgCO_3}=0,03(mol)`

`→m_C=n_C.M_C=0,03.12=0,36(g)`

Bảo toàn nguyên tố `O:`

`-n_O=3.n_{MgCO_3}=3.0,03=0,09(mol)`

`→m_O=n_O.M_O=0,09.16=1,44(g)`

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$\text{a)}$

$n_{Fe_2(SO_4)_3}=\dfrac{60}{400}=0,15\text{(mol)}$

$\to m_{Fe}=2.56.0,15=16,8\text{(g)}$

$\to m_{S}=0,15.3.32=14,4\text{(g)}$

$\to m_{O}=12.16.0,15=28,8\text{(g)}$

$\text{b)}$

$n_{NO_2}=\dfrac{3,36}{22,4}=0,15\text{(mol)}$

$\to m_{NO_2}= 0,15.46=6,9\text{(g)}$

$\to m_N= 0,15.14=2.1\text{(g)}$

$\to m_O= 0,15.2.16=4,8\text{(g)}$

$\text{c)}$

$\text{Ta có:} 1,8.10^{22}= 0,18.10^{23}$

$\to n_{MgCO_3}= \dfrac{0,18.10^{23}}{6.10^{23}}=0,03\text{(mol)}$

$\to m_{MgCO_3}=0,03.84=2,52\text{(g)}$

$\to m_{Mg}= 0,03.24=0,72\text{(g)}$

$\to m_{C}=0,03.12=0,36\text{(g)}$

$\to m_O=3.0,03.16=1,44\text{(g)}$