\(n_{Fe_2O_3}=\dfrac{20}{160}=0,125\left(mol\right)\)

PTHH:

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,125     0,375             0,125           0,375 

\(m_{ddH_2SO_4}=\dfrac{0,375.98.100}{25}=147\left(g\right)\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,125.406}{20+147}\approx30,39\%\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=\dfrac{150.3,65\%}{36,5}=0,15\left(mol\right)\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ b,m_{Zn}=0,075.65=4,875\left(g\right)\\c,m_{ddsau}=4,875+150-0,075.2=154,725\left(g\right)\\ m_{ZnCl_2}=0,075.136=10,2\left(g\right)\\c, C\%_{ddZnCl_2}=\dfrac{10,2}{154,725}.100\%\approx6,592\%\\ V_{ddsau}=V_{ddHCl}=\dfrac{150}{1,2}=125\left(ml\right)=0,125\left(l\right)\\ C_{MddZnCl_2}=\dfrac{0,075}{0,125}=0,6\left(M\right)\)

\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1..........0.3.......0.1...........0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(C\%_{HCl}=\dfrac{10.95}{150}\cdot100\%=7.3\%\)

\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_____0,1____0,3____________0,15 (mol)

b, m_HCl = 0,3.36,5 = 10,95 (g)

c, \(C\%_{HCl}=\dfrac{10,95}{150}.100\%=7,3\%\)

d, m_Al = 0,1.27 = 2,7 (g)

Bạn tham khảo nhé!

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  
          0,2       0,4           0,2 
\(m_{MgCl_2}=0,2.95=19\left(g\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

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

c, Ta có: m _{dd sau pư} = 8 + 200 = 208 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)

a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b)

n Al = 2,7/27 = 0,1(mol)

Theo PTHH :

n H2SO4 = 3/2 n Al = 0,15(mol)

=> C% H2SO4 = 0,15.98/441  .100% = 3,33%

c)

Theo PTHH :

n H2 = 3/2 n Al = 0,15(mol)

=> V H2 = 0,15.22,4 = 3,36(lít)

Sửa đề: Tính khối lượng dung dich Axit được lấy dùng cho phản ứng.

\(a)\)\(PTHH:\)

\(Mg+2CH_3COOH--->(CH_3COO)_2Mg+H_2\)

\(b)\)

\(nMg=0,3(mol)\)

Theo PTHH: \(nCH_3COOH=0,6(mol)\)

Khối lượng axit tham gia phản ứng :

\(\Rightarrow mCH_3COOH=0,6.60=36\left(g\right)\)

Khối lượng dung dich axit được lấy dùng cho phản ứng

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{36.100}{12}=300\left(g\right)_{ }\)

\(c)\)

Dung dich muối thu được sau phản ứng là \((CH_3COO)_2Mg\)

Theo PTHH: \(n(CH_3COO)_2Mg=0,3(mol)\)

\(\Rightarrow m\left(CH_3COO\right)_2Mg=42,6\left(g\right)\)

\(nH_2 =nMg=0,3(mol)\)

\(\Rightarrow mMg=7,2\left(g\right)\)

Ta có: \(m ddsau=mMg+mddCH_3COOH-mH_2\)

\(\Leftrightarrow mddsau=7,2+300-7,2=300\left(g\right)\)

Nồng độ phần trăm của dung dịch muối thu được sau phản ứng

\(C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{42,6.100}{300}=14,2\%\)

a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
b) n H2 = 3,36/22,4 = 0,15(mol)

n HCl = 2n H2 = 0,3(mol)

m HCl = 0,3.36,5 = 10,95(gam)

c) C% HCl = 10,95/150   .100% = 7,3%

d) n Al = 2/3 n H2 = 0,1(mol)

m Al = 0,1.27 = 2,7(gam)

 \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,2.174=34,8\left(g\right)\)

Ta có: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

m _{dd sau pư} = 11,2 + 200 - 0,2.2 = 210,8 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{34,8}{210,8}.100\%\approx16,51\%\)