Ta có: \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

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

_____0,2____________________0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{10}.100\%=54\%\\\%m_{Cu}=46\%\end{matrix}\right.\)

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

\(n_{HCl}=\dfrac{109,5.10\%}{36,5}=0,3\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,3}{6}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{Al\left(pư\right)}=n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 0,1.27 + 109,5 - 0,15.2 = 111,9 (g)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{111,9}.100\%\approx11,9\%\)

a, Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,4}{4}< \dfrac{0,5}{3}\), ta được: O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\)

⇒ n_{O2 (dư)} = 0,5 - 0,3 = 0,2 (mol)

⇒ m_{O2 (dư)} = 0,2.32 = 6,4 (g)

b, \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al_2O_3}=0,2.107=20,4\left(g\right)\)

a, \(2C_4H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\)

\(C_3H_8+5O_2\underrightarrow{t^o}3CO_2+4H_2O\)

b, Ta có: 58n_C4H10 + 44n_C3H8 = 160 (1)

Theo PT: \(n_{CO_2}=4n_{C_4H_{10}}+3n_{C_3H_8}=\dfrac{434}{44}\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_4H_{10}}=\\n_{C_3H_8}=\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

A là alkane → CTPT của A có dạng C_nH_2n+2

⇒ 12n + 2n + 2 = 44

⇒ n = 3

Vậy: CTPT của A là C₃H₈.

CTCT: CH₃CH₂CH₃

Q = 0,8.2877 + 0,2.2220 = 2745,6 (kJ)

Ta có: \(n_{CuSO_4}=n_{CuSO_4.5H_2O}=\dfrac{10}{250}=0,04\left(mol\right)\)

⇒ m_CuSO4 = 0,04.160 = 6,4 (g)

m _{dd CuSO4} = 10 + 90 = 100 (g)

\(\Rightarrow C\%_{CuSO_4}=\dfrac{6,4}{100}.100\%=6,4\%\)

\(xN^{+5}+\left(3x\right)e\rightarrow xN^{+2}|\times1\)

\(2yN^{+5}+\left(8y\right)e\rightarrow2yN^{+1}|\times1\)

\(Fe^{+2}\rightarrow Fe^{+3}+1e|\times\left(3x+8y\right)\)

⇒ PT: 

\(\left(3x+8y\right)FeO+\left(10x+26y\right)HNO_3\rightarrow\left(3x+8y\right)Fe\left(NO_3\right)_3+xNO+yN_2O+\left(5x+13y\right)H_2O\)

Ta có: n_HCl = 0,3.1 = 0,3 (mol)

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

_____0,1_____0,3______0,1____0,15 (mol)

⇒ x = m_Al = 0,1.27 = 2,7 (g)

y = m_AlCl3 = 0,1.133,5 = 13,35 (g)

z = V_H2 = 0,15.24,79 = 3,7185 (l)

a, PT: \(2A+6HCl\rightarrow2ACl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Theo PT: \(n_A=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{2,7}{0,1}=27\left(g/mol\right)\)

b, \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{18,35\%}=59,67\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{59,67}{1,2}=49,725\left(ml\right)\)