Question

200g H2SO4 se trateaza cu 500g KNO3 de concentratie 70%.Sa se afle care substanta este in exces si in ce cantitate.

Answer 1

H2SO4 + 2KNO3= K2SO4 + 2HNO3

n(H2SO4)=200g/98g/mol=2,04mol
500g _________100%
x____________ 70%

x=350g(KNO3)

n= 350g/101g/mol=3,47mol - exces

n(KNO3)=2mol x 2,04 mol =4,08mol
m(KNO3) = 4,08 molx101 g/mol= 412,08 mol
m(exces KNO3)= 500g-412,08g=87,92g