a) \(n_{NO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Mg + 4HNO₃ --> Mg(NO₃)₂ + 2NO₂ + 2H₂O

           0,1<--0,4<------------------------0,2

=> m_Mg = 0,1.24 = 2,4 (g)

b) \(C_M=\dfrac{0,4}{0,2}=2M\)

\(10Al+36HNO_3\rightarrow10Al\left(NO_3\right)_3+3N_2+H_2O\)

\(n_{N_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

Bảo toàn e : 

\(n_{Al}=\dfrac{10\cdot n_{N_2}}{3}=\dfrac{10}{3}\cdot0.3=1\left(mol\right)\)

\(m_{Al}=1\cdot27=27\left(g\right)\)

\(n_{HNO_3}=12n_{N_2}=12\cdot0.3=3.6\left(mol\right)\)

\(C_{M_{HNO_3}}=\dfrac{3.6}{0.1}=36\left(M\right)\)

Đáp án C

Chọn B

\(a)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

\(b)\ n_{Zn} = a\ mol\ ;\ n_{Al} = b\ mol\\ \Rightarrow65a + 27b = 6,6(1)\\ n_{H_2} = a + 1,5b = \dfrac{4,704}{22,4} = 0,21(2)\\ (1)(2)\Rightarrow a = 0,06 ; b = 0,1\\ \Rightarrow m_{Zn} = 0,06.65 = 3,9\ gam\ ;\ m_{Al} = 0,1.27 = 2,7\ gam\)

\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)

\(m_X=65a+27b=6.6\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{4.704}{22.4}=0.21\left(mol\right)\)

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

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

\(\Rightarrow a+1.5b=0.21\left(2\right)\)

\(\left(1\right),\left(2\right):\)

\(a=0.06\)

\(b=0.1\)

\(m_{Zn}=0.06\cdot65=3.9\left(g\right)\)

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

\(n_{HCl}=0,25.2=0,5\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{Zn}=0,25.65=16,25\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

Không biết đúng không nữa;-;;;

a) PTHH: Zn + 2HCl -> ZnCl2 + H2 

b) HCl=250ml=0,25l

n2HCl= V/22,4= 0,5/22,4= 0,02(mol)

Zn  +  2HCl -> ZnCl2  +  H2

1          2              1         1

0,01 <-0,5--------------> 0,01

mZn= n.M= 0,01.65= 0,65(gam)

c) VH2=n . 22,4= 0,01 . 22,4= 0,224(l)