Hoá 11, PP đường chéo

C

chonhoi110

Bài này không cần dùng pp đường chéo đâu chị à :)

Giải

$n_{P}=\dfrac{m}{31}$ (mol)

$4P+5O_2 \rightarrow 2P_2O_5$

m/31_______m/62 (mol)

$P_2O_5+H_2O \rightarrow H_3PO_4$

m/62___________m/31 (mol)

$m_{P_2O_5}=\dfrac{m}{62}.142=\dfrac{71}{31}m$ (g)

$m_{H_3PO_4}=\dfrac{m}{31}.98=\dfrac{98}{31}m$ (g)

$C\%_{H_3PO_4}=\dfrac{m_{ct}}{m_{dd}}.100\%$

$\Longleftrightarrow 9,15=\dfrac{\dfrac{98}{31}m}{\dfrac{71}{31}m+500}.100\%$

$\Longrightarrow m \approx 15,5$ (g)
 
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