\(n_{CO_2}=\dfrac{1,68}{22,4}=0,075mol\)=0,15
MgCO3+2HCl\(\rightarrow\)MgCl2+H2O+CO2(1)
R2(CO3)x+2xHCl\(\rightarrow\)2RClx+xH2O+xCO2(2)
\(\rightarrow\)\(n_{HCl}=2n_{CO_2}=0,15mol\)
\(\rightarrow\)\(m_{dd_{HCl}}=\dfrac{0,15.36,5.100}{7,3}=75gam\)
mdd sau pư=75+7,1-0,075.44=78,8gam
mMgCl2=\(\dfrac{78,8.6,028}{100}=4,75gam\)
\(\rightarrow\)\(n_{MgCl_2}=\dfrac{4,75}{95}=0,05mol\)
\(\rightarrow\)\(n_{MgCO_3}=n_{MgCl_2}=0,05mol\)\(\rightarrow\)\(m_{MgCO_3}=0,05.84=4,2gam\)
\(\rightarrow\)\(m_{R_2\left(CO_3\right)_x}=7,1-4,2=2,9gam\)
-Theo PTHH(1):\(n_{HCl\left(1\right)}=2n_{MgCl_2}=2.0,05=0,1mol\)
\(\rightarrow\)\(n_{HCl\left(2\right)}=0,15-0,1=0,05mol\)
-Theo PTHH(2):
\(n_{R_2\left(CO_3\right)_x}=\dfrac{1}{2x}n_{HCl\left(2\right)}=\dfrac{0,05}{2x}mol\)
\(\rightarrow\)\(M_{R_2\left(CO_3\right)_x}=\dfrac{2,9}{\dfrac{0,05}{2x}}=116x\)
\(\rightarrow\)2R+60x=116x\(\rightarrow\)2R=56x\(\rightarrow\)R=28x
-Nghiệm phù hợp x=2 và R=56(Fe)
