\(n_{Ba}=\frac{41,1}{137}=0,3\left(mol\right)\)
\(n_{H2SO4}=\frac{4,9\%.200}{98}=0,1\left(mol\right)\)
\(\text{Ba+2H2O}\rightarrow\text{Ba(OH)2+H2}\)
0,3____________0,3_____0,3
\(\text{Ba(OH)2+H2SO4}\rightarrow\text{BaSO4+2H2O}\)
0,3_______0,1_______0,1
\(\rightarrow\text{VH2=0,3.22,4=6,72l}\)
b) Vdd=200-0,3.2-0,1.223=176,1g
nBa(OH)2 dư= 0,3-0,1=0,2 mol
\(\rightarrow\)m Ba(OH)2 dư= 0,2.171=34,2g
\(\rightarrow\)\(C\%=\frac{\text{34,2}}{\text{176,1}}.100\%\text{=19,42%}\)
a) ba+2H2O---->Ba(OH)2+H2O
Ba(OH)2+H2SO4---->BaSO4+2H2O
n Ba=41,4/137=0,3(mol)
Theo pthh1
n H2=n Ba=0,3(mol)
V H2=0,3.22,4=6,72(l)
b)
Theo pthh1
n Ba(OH)2=n Ba=0,3(mol)
n H2SO4=\(\frac{200,4,9}{100.98}=0,1\left(mol\right)\)
--->Ba(OH)2 dư
Dd sau pư là Ba(OH)2 dư
Theo pthh2
n Ba(OH)2=n H2SO4=0,1(mol)
--->n Ba(OH)2 dư=0,2(mol)
, dd=200+0,2.171=234,2(g)
C%=\(\frac{0,2.171}{234,2}.100\%=14,6\%\)
