PTHH: CaCO3 + 2HCl \(\rightarrow\) CaCl2 + H2O + CO2\(\uparrow\) (1)
Theo pt: .... 1 ......... 2 ............ 1 ......... 1 ....... 1 .... (mol)
Theo đề: .. 0,25 .... 0,5 ........................................... (mol)
PTHH: 2HCldư + MgO \(\rightarrow\) MgCl2 + H2O (2)
Theo pt: ... 2 .......... 1 ............ 1 .......... 1 .. (mol)
Theo đề: .. 0,5 ..... 0,25 ............................. (mol)
\(n_{MgO}=\dfrac{m}{M}=\dfrac{10}{40}=0,25\left(mol\right)\)
\(m_{HCl_{dư}}=n.M=0,5.36,5=18,25\left(g\right)\)
\(m_{HCl\left(1\right)}=m_{HCl_{bđ}}-m_{HCl_{dư}}=36,5-18,25=18,25\left(g\right)\)
\(n_{HCl\left(1\right)}=\dfrac{m}{M}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
\(m_{CaCO_3}=n.M=0,25.100=25\left(g\right)\)
Vậy m = 25g