Question

Calculati masa molara medie si densitatea fata de aer respectiv fata de hidrogen a unui amestec format din 10 moli azot, 112 l (c.n) metan si 88 g CO2

Answer 1

n=10 moli N2
$n_{1}$=$\frac{112L}{22.4 \frac{L}{mol} }$=5 moli CH4
$n_{2}$=$\frac{88g}{44 \frac{g}{mol} }$=2 moli CO2
$M_{medie}$=$\frac{28*10+16*5+44*2}{10+5+2}$=$\frac{448}{17}$=26.3594
d(aer)=$\frac{ M_{medie} }{ M_{aer} }$=$\frac{26.3594}{28.9}$=0.9119
d(H2)=$\frac{ M_{medie} }{ M_{H2} }$=$\frac{26.3594}{2}$=13.1797