Ta có: \(M_{Na_2CO_3.5H_2O}=2.23+12+3.16+5.\left(2.1+16\right)=196\left(\dfrac{g}{mol}\right)\\ \rightarrow n_{Na_2CO_3.5H_2O}=\dfrac{39,2}{196}=0,2\left(mol\right)\\ \rightarrow n_{Na_2CO_3}=0,2\left(mol\right)\\ m_{HCl}=800.14,6\%=116,8\left(g\right)\\ \rightarrow n_{HCl}=\dfrac{116,8}{36,5}=3,2\left(mol\right)\\ PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\\ Tacó:\dfrac{3,2}{2}>\dfrac{0,2}{1}\)
=> HCl dư, Na2CO3 hết, tính theo nNa2CO3.
=> Chất thu dc trong dd sau phản ứng là HCl (dư) và NaCl.
nNaCl=nHCl(p.ứ)= 2. 0,2= 0,4(mol)
=> nHCl(dư) = 3,2- 0,4= 2,8(mol)
=> mNaCl= 58,5. 0,4= 23,4(g)
mHCl= 2,8. 36,5= 102,2(g)
nCO2= nNa2CO3= 0,2(mol)
=> mCO2= 0,2. 44= 8,8(g)
Theo ĐLBTKL, ta có:
mddsau= mNa2CO3.5H2O + mddHCl - mCO2
<=> mddsau= 39,2+800 - 8,8= 830,4(g)
=> C%ddHCl(dư)= (102,2/830,4).100 \(\approx\) 12,307%
C%ddNaCl= (23,4/ 830,4).100 \(\approx\) 2,818%
