250 g CaCO3 impur ............. 100%
m pur ...................................... 80%
= 200 g CaCO3 pur
a)
200 g md g V L
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
100 2x36,5 22,4
=> V = 200x22,4/100 = 44,8 L CO2 in c.n.
b)
=> md = 200x2x36,5/100 = 146 g HCl
stim ca c% = mdx100/ms
=> ms = mdx100/c%
= 146x100/20 = 730 g sol. HCl
c)
200 g n moli
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
100 1
=> n = 200x1/100 = 2 moli CO2 in c.n.
stim ca PV = nRT
P = 4 atm, T = 273+100 = 373 K
R = 0,082 , V = ? L
=. V = nRT/P
= 2x0,082x373/4 = 15,3 L CO2 (aproximativ)