Question

Help

Answer 1

14.  Đặt n _Fe2O3 =n _CuO = x

=> \(n_O=3x+x=4x\)

Ta có : \(n_O=\dfrac{3,612.10^{23}}{6,02.10^{23}}=0,6\left(mol\right)=4x\)

=> x=0,15(mol)

=> m _hh = 0,15.160 + 0,15.80 =  36(g)

=> Chọn A

Answer 2

Đặt n _SO2 =x , n _CO2 = 2x

Ta có : \(x.6,02.10^{23}+2x.6,02.10^{23}=5,418.10^{23}\)

=> x=0,3

=> \(V=\left(x+2x\right).22,4=20,16\left(lít\right)\)

=> Chọn A