\(n_{Ba}=\dfrac{41,1}{137}=0,3mol\)
\(n_{H_2SO_4}=\dfrac{200.9,8}{98.100}=0,2mol\)
Ba+H2SO4\(\rightarrow\)BaSO4+H2
-Tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\rightarrow\)Ba dư, H2SO4 hết
\(n_{Ba\left(pu\right)}=n_{BaSO_4}=n_{H_2SO_4}=0,2mol\)
\(\rightarrow\)\(n_{Ba\left(dư\right)}=0,3-0,2=0,1mol\)
Ba+2H2O\(\rightarrow\)Ba(OH)2+H2
0,1\(\rightarrow\)0,2........0.1...........0,1
\(n_{H_2}=0,2+0,1=0,3mol\)
\(V_{H_2}=0,3.22,4=6,72l\)
\(m_{dd}=41,1+200-0,2.233-0,3.2=193,9gam\)
\(m_{Ba\left(OH\right)_2}=0,1.171=17,1gam\)
C%Ba(OH)2=\(\dfrac{17,1}{193,9}.100\approx8,82\%\)
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