1.1 nFe=8,4/56=0,15mol

1.2 nO2=8/32=0,25mol

VO2=0,25*22,4=5,6l

1.3 nN2=67,2/22,4=3mol

mN2=3*28=84g

1.4 nCa=4,5.$10^{23}$ /6.$10^{23}$=0,75mol

mCa=0,75*40=30g

2.1 nCu=0,96/64=0,015mol

nBa=30,14/137=0,22mol

nCaO=70/56=1,5mol

2.4 nNH3=6,12/17=0,36mol

nBaCO3=76,83/197=0,39mol

nNH3=8,064/22,4=0,36mol

2.1 VH2=0,105*22,4=2,352l

VN2=2,24*22,4=50,176l

nNO2=3,68/46=0,08mol

VNO2=0,08*22,4=1,792l

Em tham khảo nha :

\(\begin{array}{l}
B1:\\
1)\\
{n_{Fe}} = \dfrac{{8,4}}{{56}} = 0,15mol\\
2)\\
{n_{{O_2}}} = \dfrac{8}{{32}} = 0,25mol\\
{V_{{O_2}}} = 0,25 \times 22,4 = 5,6l\\
3)\\
{n_{{N_2}}} = \dfrac{{67,2}}{{22,4}} = 3mol\\
{m_{{N_2}}} = 3 \times 28 = 84g\\
4)\\
{n_{Ca}} = \dfrac{{4,5 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,75mol\\
{m_{Ca}} = 0,75 \times 40 = 30g\\
B2:\\
1)\\
{n_{Cu}} = \dfrac{{0,96}}{{64}} = 0,015mol\\
2)\\
{n_{Ba}} = \dfrac{{30,14}}{{137}} = 0,22mol\\
3)\\
{n_{CaO}} = \dfrac{{70}}{{56}} = 1,25mol\\
4)\\
{n_{N{H_3}}} = \dfrac{{6,12}}{{17}} = 0,36mol\\
5)\\
{n_{BaC{O_3}}} = \dfrac{{76,83}}{{197}} = 0,39mol\\
6)\\
{n_{N{H_3}}} = \dfrac{{8,064}}{{22,4}} = 0,36mol\\
B3:\\
1)\\
{V_{{H_2}}} = 0,105 \times 22,4 = 2,352l\\
2)\\
{V_{{N_2}}} = 2,24 \times 22,4 = 50,176l\\
3)\\
{n_{N{O_2}}} = \dfrac{{3,68}}{{46}} = 0,08mol\\
{V_{N{O_2}}} = 0,08 \times 22,4 = 1,792l
\end{array}\)