Ta có: \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
___0,2_________0,1 (mol)
\(\Rightarrow m_{P\left(pư\right)}=0,2.31=6,2\left(g\right)\)
Mà: mP (ban đầu) = 6,89 (g)
\(\Rightarrow H\%=\dfrac{6,2}{6,89}.100\%\approx89,99\%\)
Bạn tham khảo nhé!
4P+5O2-to>2P2O5
0,2---------------0,1 mol
n P2O5=\(\dfrac{14,2}{142}\)= 0,1 mol
m P pứ= 0,2.31=6,2g
=>H %=\(\dfrac{6,2}{6,89}\)=89,98%
\(n_{P_2O_5}=\dfrac{m}{M}=\dfrac{14,2}{142}=0,1\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ Theo.PTHH:n_{P\left(pư\right)}=2.n_{P_2O_5}=2.0,1=0,2\left(mol\right)\\ m_{P\left(pư\right)}=n.M=0,2.31=6,2\left(g\right)\\ \%H=\%P=\dfrac{m_{P\left(pư\right)}}{m_{P\left(bđ\right)}}=\dfrac{6,2}{6,89}=89,98\%\)
