a)m$HNO_3$=n.M=0,75.63=47,25 (g)

b)m$CaCl_2$=n.M=0,25.111=27,75 (g)

c)n$Fe_3$$O_4$=$\frac{3.10²³}{6.10²³}$=0,5 (mol)

   m$Fe_3$$O_4$=n.M=0,5.232=116 (g)

d)n$NH_3$(đktc)=$\frac{V}{22,4}$=$\frac{3,36}{22,4}$=0,15 (mol)

   m$NH_3$=n.M=0,15.17=2,55 (g)  

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{m_{HN{O_3}}} = n \times M = 0,75 \times 63 = 47,25g\\
b)\\
{m_{CaC{l_2}}} = n \times M = 0,25 \times 111 = 27,75g\\
c)\\
{n_{F{e_3}{O_4}}} = \dfrac{A}{N} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5\,mol\\
{m_{F{e_3}{O_4}}} = n \times M = 0,5 \times 232 = 116g\\
d)\\
{n_{N{H_3}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{m_{N{H_3}}} = n \times M = 0,15 \times 17 = 2,55g
\end{array}\)