a.

\(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

b.

\(n_{BaCO_3}=n_{Na_2CO_3}=0,2.1=0,2\left(mol\right)\\ m_{kt}=197.0,2=39,4\left(g\right)\)

c.

\(n_{Ba\left(OH\right)_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{0,2.171.100\%}{200}=17,1\%\)

bạn ơi ko có CM làm sao tính được m kết tủa

đề thiếu C_m rồi

a) Na2CO3+Ba(OH)2--->BaCO3+2NaOH

b) n Na2CO3=0,1.1=0,1(mol)

Theo pthh

n BaCO3=n Na2CO3=0,1(mol)

m BaCO3=0,1.179=17,9(g)

c) Theo pthh

n Ba(OH)2=n Na2CO3=0,1(mol)

C%=\(\frac{0,1.171}{200}.100\%=8,55\%\)

C) BaCO3+2HCl---->BaCl2+H2O+CO2

Theo pthh

n HCl=2n BaCO3=0,2(mol)

m HCl=0,2.36,5=7,3(g)

a=m dd HCl=\(\frac{7,3.100}{30}=24,33\left(g\right)\)

Thiếu CM của Na2CO3 hoặc Ba(OH)2

\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)

PTHH :

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

0,2              0,4                  0,2               0,2 

\(m_{NaOH}=0,4.40=16\left(g\right)\)

\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)

\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)

\(d,PTHH:\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

0,2               0,2 

\(m_{CuO}=0,2.80=16\left(g\right)\)

a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)

c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)

d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

\(a)Ba\left(OH\right)_2+Na_2CO_3\rightarrow2NaOH+BaCO_3\\ n_{Ba\left(OH\right)_2}=0,2.2=0,4mol\\ n_{BaCO_3}=n_{Na_2CO_3}=n_{Ba\left(OH\right)_2}=0,4mol\\ m_{BaCO_3}=0,4.171=68,4g\\ b)V_{Na_2CO_3}=\dfrac{0,4}{1}=0,4l\\ c)n_{NaOH}=2n_{Ba\left(OH\right)_2}=0,8mol\\ C_{M\left(NaOH\right)}=\dfrac{0,8}{0,2+0,4}=\dfrac{4}{3}M\)

PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5\left(mol\right)=n_{BaSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,5\cdot98}{15\%}\approx326,7\left(g\right)\\m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\end{matrix}\right.\)

\(n_{Ba\left(OH\right)_2}=0,5\cdot1=0,5mol\)

a)\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

    0,5                 0,5            0,5        

b) \(m_{H_2SO_4}=0,5\cdot98=49\left(g\right)\)

   \(\Rightarrow m_{ddH_2SO_4}=\dfrac{49}{15}\cdot100=326,67\left(g\right)\)

c) \(m_{BaSO_4}=0,5\cdot233=116,5\left(g\right)\)