DE CUONG ON TAP HK1 MON HOA HOC
1. hoa tan 0,54 gam nhom vao dd 120 gam dd h2so4 4,9% thoat ra V lit khi hidro (dktc)
a) viet pthh . tinh V ?
b) tinh nong do phan tram cua cac chat trong dd sau phan ung ?
2. hoa tan 15,5 gam na2o vao nuoc thanh 500 ml dung dich A
a) viet pthh xay ra
b) tinh nong do mol cua dd A
c) tinh the tich dd h2so4 20% ( D = 1,14g/ml ) can de trung hoa luong dd tren
1.
Theo đề bài ta có : \(\left\{{}\begin{matrix}nAl=\dfrac{0,54}{27}=0,02\left(mol\right)\\nH2SO4=\dfrac{120.4,9}{100.98}=0,06\left(mol\right)\end{matrix}\right.\)
PTHH :
\(2Al+3H2SO4->Al2\left(So4\right)3+3H2\uparrow\)
0,02mol...0,03mol.......0,01mol.............0,03mol
Theo PTHH ta có : nAl = \(\dfrac{0,02}{2}mol< nH2SO4=\dfrac{0,06}{2}mol=>nH2SO4\left(dư\right)\) ( tính theo nal)
=> VH2(đktc) = 0,03.22,4 = 6,72(l)
=> \(\left\{{}\begin{matrix}C\%ddH2SO4\left(dư\right)=\dfrac{\left(0,06-0,03\right).98}{0,54+120-0,03.2}.100\%\approx2,44\%\\C\%ddAl2\left(SO4\right)3=\dfrac{0,01.302}{0,54+120-0,03.2}.100\%\approx2,5\%\end{matrix}\right.\)
Theo đề bài ta có : nNa2O = \(\dfrac{15,5}{62}=0,25\left(mol\right)\)
a) PTHH :
\(Na2O+H2O->2NaOH\)
0,25mol....0,25mol.....0,5mol
b) Nồng độ mol dd A là :
CMddNaOH = 0,5/0,5 = 1(M)
c) PTHH :
\(2NaOH+H2SO4->Na2SO4+H2O\)
0,5mol.........0,25mol
=> mddH2SO4 = \(\dfrac{0,25.98}{20}.100=122,5\left(g\right)=>VddH2SO4=\dfrac{122,5}{1,14}\approx107,5\left(ml\right)\)