Em tham khảo nha:

\(\begin{array}{l}
2)\\
a)\,BaC{l_2} + N{a_2}S{O_4} \to BaS{O_4} + 2NaCl\\
b)\\
{n_{BaC{l_2}}} = \dfrac{{200 \times 10,4\% }}{{208}} = 0,1\,mol\\
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,1\,mol\\
{m_{BaS{O_4}}} = 0,1 \times 233 = 23,3g\\
c)\\
{m_{{\rm{dd}}spu}} = 200 + 400 - 23,3 = 576,7g\\
{n_{NaCl}} = 2{n_{BaC{l_2}}} = 0,2\,mol\\
{C_\% }NaCl = \dfrac{{0,2 \times 58,5}}{{576,7}} \times 100\%  = 2,03\% \\
3)\\
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
b)\\
{n_{N{a_2}C{O_3}}} = \dfrac{{10,6}}{{106}} = 0,1\,mol\\
{n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,2\,mol\\
{C_\% }HCl = \dfrac{{0,2 \times 36,5}}{{250}} \times 100\%  = 2,92\% \\
c)\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,1\,mol\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,2\,mol\\
{m_{{\rm{dd}}spu}} = 10,6 + 250 - 0,1 \times 44 = 351,6g\\
{C_\% }NaCl = \dfrac{{0,2 \times 58,5}}{{351,6}} \times 100\%  = 3,33\% 
\end{array}\)

 

Đáp án:

Bài 2 : b) 23.3 gam

c) 2.03%

Bài 3 

b) 

 

Giải thích các bước giải:

Bài 2 : 

mBaCl2 = 200 × 10.4% = 20.8 gam

nBaCl2 = 20.8 ÷ 208 = 0.1 mol

BaCl2 + Na2SO4 --> BaSO4 + 2NaCl 

0.1                                 0.1      0.2  (mol)

Theo PTHH : nBaSO4 = nBaCl2= 0.1 mol

Theo pthh : nNaCl = 2nBaCl2 = 0.2 mol

a ) BaCl2 + Na2SO4 ---> BaSO4 + NaCl

b ) mBaSO4 = 0.1 × 233 = 23.3 gam

c ) mNaCl = 0.2 × 58.5 = 11.7 gam

mdd sau pư = 200 + 400 - 23.3 = 576.7 g

C%NaCl = (11.7 ÷ 576.7 ) × 100 = 2.03%

Bài 3 : 

nNa2CO3 = 10.6 ÷ 106 = 0.1 mol

Na2CO3 + 2HCl ---> 2NaCl + CO2 + H2O 

0.1mol    0.2mol.    0.2 mol. 0.1 mol

Theo pthh : nHCl = 2nNa2CO3 = 0.2 mol

nNaCl = 2nNa2CO3 = 0.2 mol

nCO2 = nNa2CO3 = 0.1 mol

a) Na2CO3 + 2HCl --->2NaCl+CO2 + H2O

b) mHCl = 0.2 × 36.5 = 7.3 g

C%HCl = (7.3 ÷ 250) × 100 = 2.92%

c) mNaCl = 0.2 × 58.5 = 11.7 g

mdd sau pư=10.6+250 -0.1×44 = 256.2 g

C%NaCl = (11.7 ÷ 256.2)× 100 = 4.567%