C₆H₆+HNO₃ → C₆H₅NO₂+ H₂O
a) ρ = [tex]\frac{m}{V} =>[/tex] ρ * V = m (de pe ecuatia reactiei chimice tot timpul scoatem md) => m = [tex]\frac{0.88g}{ml} * 100 l = \frac{0.88g}{ml} * 100*10^{3}ml = 88000g=88kg C6H6[/tex]
M(C₆H₆) = 6 * A(C) + 6* A(H) = 6 * 12 + 6 * 1 = 72 + 6 = 78
M(HNO₃) = 1 * A(H) + 1 * A(N) + 3 * A(O) = 1 * 1 + 1* 14 + 3 * 16 = 1 + 14 + 48 = 63
[tex]\frac{x}{63kg}=\frac{88kg}{78kg} => x = \frac{88*63kg}{78} => x = 71.07kgHNO3[/tex]
[tex]C=\frac{md*100}{ms} => ms = \frac{md*100}{C} => ms = \frac{71,07kg*100}{63} => ms =112.8 kg HNO3[/tex]
b) η=[tex]\frac{mp*100}{mt} => mp = \frac{randement*mt}{100}[/tex]
aflam masa teoretica de C₆H₅NO₂ (de pe ecuatia reactiei chimice, in conditiile in care tot benzenul reactioneaza cu tot acidul.
M(C₆H₅NO₂)= 6 * A(C) + 5* A(H) + 1 * A(N) + 2 * A(O) = 6 * 12 + 5 * 1 + 1* 14 + 2 * 16 = 72+5+14+32=123
[tex]\frac{y}{123kg} = \frac{88kg}{78kg} => y=\frac{123*88kg}{78} => y=138,76kg C6H5NO2[/tex]
[tex]mp = \frac{70*138,76kg}{100} => mp=97,13 kg C6H5NO2[/tex]
