Ta có: \(4.M_C=3.M_X\Leftrightarrow M_X=\dfrac{4.12}{3}=16\left(g/mol\right)\)

    ⇒ X là oxi

a, 

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:       x             x                 x

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂

Mol:        y              3y                    y

Ta có: \(\left\{{}\begin{matrix}80x+160y=56\\160x+400y=136\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,3\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right);m_{Fe_2O_3}=56-8=48\left(g\right)\)

b, \(\%m_{CuO}=\dfrac{8.100\%}{56}=14,29\%\)

\(\%m_{Fe_2O_3}=\dfrac{48.100\%}{56}=85,71\%\)

c, \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,1+3.0,3=1\left(mol\right)\)

\(\Rightarrow C_{M_{ddH_2SO_4}}=\dfrac{1}{0,2}=5M\)

Ta có: \(\left\{{}\begin{matrix}p+e+n=48\\p=e\\e=25\%\left(p+e+n\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=12\\n=24\end{matrix}\right.\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + H₂SO4 → MgSO₄ + H₂

Mol:      0,2                                    0,2

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

 ⇒ Chọn C

Đổi 54km/h = 15m/s; 1p = 60s

Gia tốc của tàu là:

Ta có: \(v=v_0+at\Leftrightarrow a=\dfrac{v-v_0}{t}=\dfrac{0-15}{60}=-0,25\left(m/s^2\right)\)

Ta có: \(\left\{{}\begin{matrix}p+e-n=14\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=15\\p=e\\n-p=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}p=e=15\\n=16\end{matrix}\right.\)