PTHH: \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15mol\\n_{Al_2\left(SO_4\right)_3}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\end{matrix}\right.\)

\(a,n_{\left(NH_4\right)_3PO_4}=0,6\left(mol\right)\\ \Rightarrow n_N=0,6.3=1,8\left(mol\right)\Rightarrow m_N=1,8.14=25,2\left(g\right)\\ n_H=4.3.0,6=7,2\left(mol\right)\Rightarrow m_H=7,2.1=7,2\left(g\right)\\ n_P=n_{hc}=0,6\left(mol\right)\Rightarrow m_P=0,6.31=18,6\left(g\right)\\ n_O=4.0,6=2,4\left(mol\right)\Rightarrow m_O=2,4.16=38,4\left(g\right)\)

\(b,n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=342.\dfrac{1}{15}=22,8\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{20,52}{342}=0,06\left(mol\right)\\ n_O=4.3.0,06=0,72\left(mol\right)\\ \Rightarrow n_{CO_2}=\dfrac{0,72}{2}=0,36\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right)\)

Bài 1: Tính Khối Lượng Của Nguyên Tố Oxi có trong mỗi hợp chất sau:
1. 18 gam nước

\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\Rightarrow n_O=1.1=1\left(mol\right)\)

=> m_O = 1.16 = 16 (g)
2. 2,2 gam CO₂

\(n_{CO_2}=\dfrac{2,2}{44}=0,05\left(mol\right)\Rightarrow n_O=0,05.2=0,1\left(mol\right)\)

m_O = 0,1 .16 =1x6(g)

3. 8 gam CuSO₄

\(n_{CuSO_4}=\dfrac{8}{160}=0,05\left(mol\right)\Rightarrow n_O=0,05.4=0,2\left(mol\right)\)

=> m_O= 0,2.16= 3,2(g)

4. 2 gam Fe₂(SO₄)₃

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{2}{400}=0,005\left(mol\right)\Rightarrow n_O=0,005.12=0,06\left(mol\right)\)

=> m_O= 0,06.16 = 0,96(g)

Bài 1 : 

Số mol , khối lượng , số phân tử của các chất lần lượt là : 

\(a.\)\(\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(m_{O_2}=0.05\cdot32=1.6\left(g\right)\)

\(0.05\cdot6\cdot10^{23}=0.3\cdot10^{23}\left(pt\right)\)

\(b.\)

\(n_{SO_3}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(m_{SO_3}=0.1\cdot80=8\left(g\right)\)

\(0.1\cdot6\cdot10^{23}=0.6\cdot10^{23}\left(pt\right)\)

\(c.\)

\(n_{H_2S}=\dfrac{36}{22.4}=\dfrac{45}{28}\left(mol\right)\)

\(m_{H_2S}=\dfrac{45}{28}\cdot34=\dfrac{765}{14}\left(g\right)\)

\(\dfrac{45}{28}\cdot6\cdot10^{23}=\dfrac{135}{14}\cdot10^{23}\left(pt\right)\)

\(d.\)

\(n_{C_4H_{10}}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(m_{C_4H_{10}}=0.2\cdot58=11.6\left(g\right)\)

\(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

Bài 2 : 

\(a.\)

\(n_{SO_3}=\dfrac{16}{80}=0.2\left(mol\right)\)

Số phân tử SO₃ : \(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

\(b.\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

Số phân tử NaOH : \(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

\(c.\)

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{16}{400}=0.04\left(mol\right)\)

Số phân tử Fe₂(SO₄)₃ : \(0.04\cdot6\cdot10^{23}=0.24\cdot10^{23}\left(pt\right)\)

\(d.\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{34.2}{342}=0.1\left(mol\right)\)

Số phân tử Al₂(SO₄)₃ : \(0.1\cdot6\cdot10^{23}=0.6\cdot10^{23}\left(pt\right)\)

Bài 2 Tính số mol , số phân tử các chất sau:

a, 16 gam SO3

\(n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\)

b, 8 gam NaOH

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

c, 16 gam Fe2(SO4)3

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{16}{400}=0,04\left(mol\right)\)

d, 34,2 gam Al2(SO4)3

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{342}=0,1\left(mol\right)\)

a)

$n_O = n_{NaOH} = \dfrac{12}{40} = 0,3(mol)$
$m_O = 0,3.16 = 4,8(gam)$

b)

$n_{Fe_2(SO_4)_3} = \dfrac{40}{400} = 0,1(mol)$
$n_O = 12n_{Fe_2(SO_4)_3} = 1,2(mol)$
$m_O = 1,2.16 = 19,2(gam)$

c)

$n_{Na_3PO_4} = \dfrac{8,2}{164} = 0,05(mol)$
$n_O = 4n_{Na_3PO_4} = 0,05.4 = 0,2(mol)$
$m_O = 0,2.16 = 3,2(gam)$

\(n_{Al2O3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Pt : \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

          1              3                    1                3

        0,2           0,6                   0,2

\(n_{H2SO4}=\dfrac{0,2.3}{1}=0,6\left(mol\right)\)

⇒ \(m_{H2SO4}=0,6.98=58,8\left(g\right)\)

\(n_{Al2\left(SO4\right)3}=\dfrac{0,6.1}{3}=0,2\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,2.342=68,4\left(g\right)\)

 Chúc bạn học tốt

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ \rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)

Câu 1:

Ta có: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{61,56}{342}=0,18\left(mol\right)\)

PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

____0,18_______________0,18 (mol)

\(\Rightarrow m_{Al_2O_3}=0,18.102=18,36\left(g\right)\)

Câu 2:

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Xét tỉ lệ:\(\dfrac{0,2}{2}< \dfrac{0,15}{1}\), ta được O₂ dư.

Theo PT: \(n_{MgO}=n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

Bạn tham khảo nhé!

Câu 3:

Ta có: \(n_{CaCO_3}=\dfrac{72}{100}=0,72\left(mol\right)\)

Vì: H% = 80% \(\Rightarrow n_{CaCO_3\left(pư\right)}=0,72.80\%=0,576\left(mol\right)\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

____0,576__0,576 (mol)

\(\Rightarrow m_{CaO}=0,576.56=32,256\left(g\right)\)

Câu 4:

Quy đổi hỗn hợp về FeO và Fe₂O₃.

PT: \(FeO+2HCl\rightarrow FeCl_2+H_2O\) (1)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\) (2)

Có: \(n_{FeCl_2}=\dfrac{20,32}{127}=0,16\left(mol\right)=n_{FeO}\)

\(\Rightarrow n_{HCl\left(1\right)}=2n_{FeCl_2}=0,32\left(mol\right)\)

Mà: \(n_{HCl}=\dfrac{46,72}{36,5}=1,28\left(mol\right)\Rightarrow n_{HCl\left(2\right)}=0,96\left(mol\right)\)

\(\Rightarrow n_{Fe_2O_3}=\dfrac{1}{6}n_{HCl\left(2\right)}=0,16\left(mol\right)\)

⇒ m = m_FeO + m_Fe2O3 = 0,16.72 + 0,16.160 = 37,12 (g)

Bạn tham khảo nhé!

Câu a.

\(M_{Ca\left(NO_3\right)_2}=164\)g/mol

\(m_{Ca\left(NO_3\right)_2}=0,3\cdot164=49,2g\)

\(\%Ca=\dfrac{40}{164}\cdot100\%=24,39\%\)

\(m_{Ca}=\%Ca\cdot49,2=12g\)

\(\%N=\dfrac{14\cdot2}{164}\cdot100\%=17,07\%\)

\(m_N=\%N\cdot49,2=8,4g\)

\(m_O=49,2-12-8,4=28,8g\)

Các câu sau em làm tương tự nhé!