Question

ofer 15p
Nr de moli de CaO continuti in 500g oxid de calciu de puritate 90 la suta

Answer 1

$m_{CaO}=500g$ (masa impura)
$P=90$%

$P=\frac{m_{pur}}{m_{impur}}*100=\ \textgreater \ m{pur}=\frac{P*m_{impur}}{100}=\frac{90*500}{100}=90*5=450g$ CaO

$M_{CaO}=40+16=56g/mol$

$n=\frac{m(pur)}M={450}{56}=8.03$ moli CaO