Question

Answer 1

Câu 2 : 

\(SO_3+H_2O\rightarrow H_2SO_4\) : P/ư hóa hợp

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) : P/ư thế

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\) : P/ư hóa hợp

\(Na_2O+H_2O\rightarrow2NaOH\) : P/ư hóa hợp

Answer 2

Câu 3 : 

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2.......0.4...........0.2.........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0.4}{0.2}=2\left(M\right)\)

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

\(\dfrac{0.4}{3}......................0.2\)

\(m_{Al}=\dfrac{0.4}{3}\cdot27=3.6\left(g\right)\)

 

Answer 3

Câu 4 : 

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

\(0.2.......................0.2..........0.1\)

\(m_{\text{dung dịch sau phản ứng}}=4.6+95.6-0.1\cdot2=100\left(g\right)\)

\(C\%_{NaOH}=\dfrac{0.2\cdot40}{100}\cdot100\%=8\%\)

Answer 4

Câu 4 : 

n Na = 4,6/23 = 0,2(mol)

$2Na + 2H_2O \to 2NaOH + H_2$

n H2 = 1/2 n Na = 0,1(mol)

m dd = m Na + m H2O - m H2 = 4,6 + 95,6 - 0,1.2 = 100(gam)

C% NaOH = 0,2.40/100 .100% = 8%