Ta co pthh
2Na + H2O \(\rightarrow\)Na2O+H2
Ba + H2O \(\rightarrow\)BaO + H2
Theo de bai ta co
nH2O=\(\dfrac{59,725}{18}\approx3,3mol\)
nH2=\(\dfrac{10,08}{22,4}=0,45mol\)
theo 2 pthh
nH2O=\(\dfrac{3,3}{2}mol>nH2=\dfrac{0,45}{2}mol\)
\(\Rightarrow\)so mol cua H2O du ( tinh theo so mol cua H2 )
Goi x la so mol cua H2 tham gia vao pthh 1
so mol cua H2 tham gia vao pthh 2 la 0,45-x mol
theo pthh
nNa=2nH2=2x mol
nBa=nH2=0,45-x mol
theo de bai ta co he phuong trinh
23*2x+137*(0,45-x)=41,175
\(\Leftrightarrow\)46x+61,65-137x=41,175
\(\Leftrightarrow\)46x-137x=41,175-61,65
\(\Leftrightarrow\)-91x=-20,475
\(\Rightarrow\)x= \(\dfrac{-20,475}{-91}=0,225mol\)
\(\Rightarrow\)so mol cua H2 tham gia vao pthh 2 la 0,45-0,225=0.225mol
Theo 2 pthh
nH2O= nH2=0,225mol
nNa2O=nH2=0,225 mol
nBaO=nH2=0,225 mol
\(\Rightarrow\)mH2O=0,225*18=4,05g
mNa2O=0,225*62=13,95 g
mBaO=0,225*153=34,425 g
\(\Rightarrow\)Nong do phan tram cua cac chat trong dd X la
C% cua Na2O = \(\dfrac{mct}{mdd}100\%\)=\(\dfrac{13,95}{\left(13,95+4,05\right)}100\%=77,5\%\)
C% cua BaO=100%-77,5%=22,5%
