mFe=n.M=0.1*56=5.6 g

mH2O=n.M=0.25*18=4.5 g

nAl=m/M=5.4/27=0.2 mol

nNaCl=m/M=58.5/58.5=1 mol

nH2=V/22.4=2.4/22.4=0.107 mol

nNH3=V/22.4=5.6/22.4=0.205 mol

#hoctot

xin câu trả lời hay nhất !! 

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{m_{Fe}} = n \times M = 0,1 \times 56 = 5,6g\\
{m_{{H_2}O}} = n \times M = 0,25 \times 18 = 4,5g\\
b)\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
{n_{NaCl}} = \dfrac{m}{M} = \dfrac{{58,5}}{{58,5}} = 1\,mol\\
c)\\
{n_{N{H_3}}} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol
\end{array}\)

Em xem lại chỗ 2,4 H2 nha