dot chay het 6.2 g photpho trong khong khi thu duoc di photpho pentaoxit
a) tinh k.luong p2o5 thu duoc
b) tinh the tich khong khi o dktc can dung biet rang the tich khong tich oxi chiem 1/5 the tich khong khi
dot chay het 6.2 g photpho trong khong khi thu duoc di photpho pentaoxit
a) tinh k.luong p2o5 thu duoc
b) tinh the tich khong khi o dktc can dung biet rang the tich khong tich oxi chiem 1/5 the tich khong khi
$n_P=6,2/31=0,2mol$
$4P+5O_2\overset{t^o}\to 2P_2O_5$
$\text{a/Theo pt :}$
$n_{P_2O_5}=1/2.n_P=1/2.0,2=0,1mol$
$⇒m_{P_2O_5}=0,1.142=14,2g$
$\text{b/Theo pt :}$
$n_{O_2}=5/4.n_P=5/4.0,2=0,25mol$
$⇒V_{kk}=0,25.22,4.5=28l$
$Số$ $mol$ $của$ $6,2(g)$ $P$ $là:$
$n_{P}=\frac{m_{P}}{M_{P}}=\frac{6,2}{31}=0,2(mol)$
$PTHH:4P+5O_{2}→t^{o}→2P_{2}O_{5}$
$TPT:$$4$ $5$ $2$$(mol)$
$TĐB:$$0,2$ $0,25$ $0,1$$(mol)$
$a)$$Khối$ $lượng$ $của$ $0,1(mol)$ $P_{2}O_{5}$ $là:$
$m_{P_{2}O_{5}}=n_{P_{2}O_{5}}.M_{P_{2}O_{5}}=0,1.142=14,2(g)$
$b)$$Thể$ $tích$ $của$ $0,25(mol)$ $O_{2}$ $là:$
$V_{O_{2}}=n_{O_{2}}.22,4=0,25.22,4=5,6(l)$
$Do$ $thể$ $tích$ $không$ $khí$ $gấp$ $5$ $lần$ $thể$ $tích$ $oxi$
$\Rightarrow V_{kk}=5V_{O_{2}}=5.5,6=28(l)$