$a) n_{Mg} = \dfrac{7,2}{24} = 0,3 (mol)$

PTHH: $Mg + 2HCl \rightarrow MgCl_2 + H_2$

            0,3--------------------------------->0,3

$b) \rightarrow V_{H_2} = 0,3.24,79 = 7,437 (l)$

$c)$ PTHH: $Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

                                       0,15----------------->0,1

$\rightarrow m_{Fe} = 0,1.56 = 5,6 (g)$

Đáp án:

 

Giải thích các bước giải:

a)

\(Mg + 2HCl \to MgCl_2^{} + H_2^{}\)

b)

\(n_{Mg}^{} = \frac{m}{M} = \frac{{7,2}}{{24}} = 0,3{\text{ mol}}\)

Theo PTHH thì:

\({\text{n}}_{H_2^{}}^{} = n_{Mg}^{} = 0,3{\text{ mol}}\)

\({\text{V}}_{H_2^{}}^{} = n_{H_2^{}}^{}.22,4 = 0,3.22,4 = 6,72l\)

c)

\(Fe_2^{}O_3^{} + 3H_2^{}\xrightarrow{{t_{}^o}}2Fe + 3H_2^{}O\)

Theo PTHH thì:

\(n_{Fe}^{} = \frac{2}{3}n_{H_2^{}}^{} = 0,2{\text{ mol}}\)

\({\text{m}}_{Fe}^{} = 0,2.56 = 11.2g\)