\(n_{N_2}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}=0.2\left(mol\right)\)
\(n_{CO}=\dfrac{1.8\cdot10^{23}}{6\cdot10^{23}}=0.3\left(mol\right)\)
\(m_{hh}=0.2\cdot28+0.3\cdot28=14\left(g\right)\)
Ta có: nN2= \(\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\)
nCO= \(\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\)
mN2= 0,2.28= 5,6(g)
mCO= 0,3.28 = 8,4(g)
=> \(m_{dd}\)= 5,6 + 8,4 = 14(g)
Vậy đáp án đúng là B