\(n_{K_2CO_3}=0.1\cdot0.5=0.05\left(mol\right)\)

\(n_{CaCl_2}=0.1\cdot0.1=0.01\left(mol\right)\)

\(K_2CO_3+CaCl_2\rightarrow CaCO_3+2KCl\)

Lập tỉ lệ : 

\(\dfrac{0.05}{1}>\dfrac{0.01}{1}\) \(\Rightarrow K_2CO_3dư\)

\(n_{CaCO_3}=n_{CaCl_2}=0.01\left(mol\right)\)

\(m=0.01\cdot100=1\left(g\right)\)

\(b.\)

Các chất có trong dung dịch : 

\(K_2CO_3\left(dư\right):0.04\left(mol\right),KCl:0.02\left(mol\right)\)

\(V=0.1+0.1=0.2\left(l\right)\)

\(\left[K^+\right]=\dfrac{0.04\cdot2+0.02}{0.2}=0.5\left(M\right)\)

\(\left[CO_3^{2-}\right]=\dfrac{0.04}{0.2}=0.2\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.02}{0.2}=0.1\left(M\right)\)

PTHH: \(Na_2CO_3+Ca\left(OH\right)_2\rightarrow2NaOH+CaCO_3\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=0,1\cdot1=0,1\left(mol\right)\\n_{Ca\left(OH\right)_2}=0,1\cdot1,5=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Ca(OH)₂ dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CaCO_3}=0,1\left(mol\right)\\n_{Ca\left(OH\right)_2\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,1\cdot100=10\left(g\right)\\C_{M_{NaOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\\C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\end{matrix}\right.\)

nNa2CO3= 0,1(mol) ; nCa(OH)2=0,15(mol)

a) PTHH: Na2CO3 + Ca(OH)2 -> CaCO3 + 2 NaOH

Vì: 0,1/1 < 0,15/1

=> Na2CO3 hết, Ca(OH)2 dư, tính theo Na2CO3.

=> nCaCO3=nCa(OH)2 (p.ứ)=nNa2CO3= 0,1(mol)

=>m(kết tủa)=mCaCO3=0,1.100=10(g)

b) Vddsau= 100+100=200(ml)=0,2(l)

nNaOH=2.0,1=0,2(mol)

nCa(OH)2(dư)=0,15-0,1=0,05(mol)

=>CMddNaOH=0,2/0,2= 1(M)

CMddCa(OH)2 (dư)= 0,05/ 0,2=0,25(M)

\(Na_2CO_3+CaCl_2\rightarrow2NaCl+CaCO_3\left(1\right)\)

\(n_{Na_2CO_3}=\dfrac{50}{1000}.0,2=0,01\left(mol\right)\)

\(n_{CaCl_2}=\dfrac{100}{1000}.0,15=0,015\left(mol\right)\)

Theo (1), ta có:

\(\dfrac{0,01}{1}< \dfrac{0,015}{1}\)\(\Rightarrow Na_2CO_3\) hết, \(CaCl_2\) dư, vậy tính kết tủa theo Na₂CO₃.

\(n_{CaCO_3}=n_{Na_2CO_3}=0,01\left(mol\right)\)

Khối lượng kết tủa: \(m_{CaCO_3}=0,01.100=1\left(g\right)\)

\(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\left(2\right)\)

Theo đề cho thì khối lượng kết tủa BaCO₃ bằng khối lượng kết tủa CaCO₃ bằng 1(g)

\(\Rightarrow n_{BaCl_2}=n_{BaCO_3}=\dfrac{1}{197}\left(mol\right)\)

\(a=C_{M\left(BaCl_2\right)}=\dfrac{\dfrac{1}{197}}{\dfrac{100}{1000}}\approx0,05\left(M\right)\)

PTHH: \(Na_2SO_4+CaCl_2\rightarrow2NaCl+CaSO_4\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\cdot0,5=0,05\left(mol\right)\\n_{CaCl_2}=0,1\cdot0,4=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Na₂SO₄ dư

\(\Rightarrow\left\{{}\begin{matrix}n_{CaSO_4}=0,04\left(mol\right)\\n_{NaCl}=0,08\left(mol\right)\\n_{Na_2SO_4\left(dư\right)}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_4}=0,04\cdot136=5,44\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,08}{0,1+0,1}=0,4\left(M\right)\\C_{M_{Na_2SO_4\left(dư\right)}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)

Na2CO3+BaCl2---->2NaCl+BaCO3

Ta có

n Na2CO3=0,5.0,1=0,05(mol)

n BaCl2=0,1.1=0,1(mol)

---->BaCl2 dư..

Theo pthh

n BaCO3=n Na2CO3=0,05(mol)

m BaCO3=0,05.`197=9,85(g)

CaCl₂ trộn với NaOH không tạo kết tủa nha em!