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

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

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

Gọi CTHH cần tìm là C_xH_yN_t

%N = 100 - 28,75 - 4,76 = 66,49%

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{126.28,75\%}{12}=3\\y=\dfrac{126.4,76\%}{1}=6\\t=\dfrac{126.66,49\%}{14}=6\end{matrix}\right.\)

Vậy: CTHH cần tìm là C₃H₆N₆.

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

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

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PT: n_H2SO4 = n_H2 = 0,1 (mol)

Thep ĐLBT KL: m_KL + m_H2SO4 = m _muối + m_H2

⇒ m _muối = 10 + 0,1.98 - 0,1.2 = 19,6 (g)

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

Ta có: m_{CaCO3 (pư)} = m_CaO + m_CO2 = 224 + 176 = 400 (g)

\(\Rightarrow H\%=\dfrac{400}{500}.100\%=80\%\)

a, V_H2O = 250:1 = 250 (ml) = 0,25 (l)

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

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

a, \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

b, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,075\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,075.24,79=1,85925\left(l\right)\)

Ta có: \(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

PT: \(Mg+CuSO_4\rightarrow MgSO_4+Cu\)

_____0,2_______0,2___________0,2 (mol)

a, m_Mg = 0,2.24 = 4,8 (g)

b, \(C\%_{CuSO_4}=\dfrac{0,2.160}{150}.100\%\approx21,33\%\)