Đố biết

H₂O (Nước)

\(n_{Na_2CO_3}=\dfrac{12,72}{106}=0,12\left(mol\right)\)

PT: \(Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\)

Theo PT: n_CaCO3 = n_Na2CO3 = 0,12 (mol)

⇒ m = m_CaCO3 = 0,12.100 = 12 (g)

Có lẽ đề cho khí H₂ ở đktc chứ không phải đkc bạn nhỉ?

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m _{dd sau pư} = 0,2.27 + 300 - 0,3.2 = 304,8 (g)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{304,8}.100\%\approx11,2\%\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Theo PT: \(n_{Al_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)

Mà: H% = 60%

\(\Rightarrow n_{Al_2O_3\left(TT\right)}=0,1.60\%=0,06\left(mol\right)\)

⇒ m_{Al2O3 (TT)} = 0,06.102 = 6,12 (g)

a, \(2Mg+O_2\underrightarrow{t^o}2MgO\)

b, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

⇒ m_MgO = 0,4.40 = 16 (g)

c, \(n_{MgO}=\dfrac{20}{40}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,25\left(mol\right)\)

⇒ V_O2 = 0,25.22,4 = 5,6 (l)

Ta có: n_Ca(OH)2 = 0,1.2 = 0,2 (mol)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

_____0,2________0,2_______0,2 (mol)

a, V = 0,2.24,79 = 4,958 (l)

b, m_CaCO3 = 0,2.100 = 20 (g) 

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

b, Ta có: \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,75\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,75.24,79=18,5925\left(l\right)\)

c, \(n_{HCl}=3n_{Al}=1,5\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{1,5}{1}=1,5\left(l\right)=1500\left(ml\right)\)

d, \(n_{AlCl_3}=n_{Al}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{AlCl_3}}=\dfrac{0,5}{1,5}=\dfrac{1}{3}\left(M\right)\)

chịu

tui hổng bt

bó tay