Hóa 9

F

furelove

Gọi 98% $ m_H_3PO_4=a$
98% $m_H_2SO_4=b$
m H trong a g$ H_3PO_4= \frac{3}{98}\ a$
m H trong b g$ H_2SO_4=\frac{2}{98}b$
mà theo đề: m H trong a g $H_3PO_4=$ m H trong b g $H_2SO_4$
\Rightarrow $ \frac{3}{98}a=\frac{2}{98}b$\Leftrightarrow 3a=2b
Ta có: m O trong a g $H_3PO_4$=$\frac{16*4}{98}=\frac{32}{49} a$
m S trong b g $H_2SO_4$=$\frac{32*4}{98}=\frac{64}{49} b$
\Rightarrow m O: m S=$ \frac{32a}{49}:\frac{64b}{49}=\frac{a}{2b}=\frac{a}{3a}=1/3$
 
You must log in or register to reply here.
Top Bottom