hóa học 10

L

lp_qt

$n_{HCl}=\dfrac{m}{10.36,5}$

$KL + 2HCl \rightarrow MuốiCl + H_2$

áp dụng ĐLBTKL:

$6.05+\dfrac{m}{10}=13.15+\dfrac{m}{10.36,5}$

\Leftrightarrow $m=73$
 
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