Question

cho m g CaCO3 tac dung voi dung dich chua 36,5g axit HCl luong axit du pu vua du vs 10g MgO tinh m

Answer 1

CHÚC BẠN HỌC TỐT!!

Theo đề bài, ta có: \(\left\{{}\begin{matrix}n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\\n_{MgO}=\dfrac{10}{40}=0,25\left(mol\right)\end{matrix}\right.\)

PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2\)

pư..........0,25..........0,5...............0,25.......0,25 (mol)

\(\Rightarrow n_{HClpư}=1-0,5=0,5\left(mol\right)\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

pư...............0,25..........0,5...........0,25........0,25...........0,25 (mol)

\(\Rightarrow m_{CaCO3}=100.0,25=25\left(g\right)\)

Vậy...............