Question

Tính% khối lượng mỗi nguyên tố trong hợp chất.

3H₃PO_{4 ,} 10Ca(OH)

 

Giúp m với:)

Answer 1

\(M_{3H_3PO_4}=\left(3.1.3\right)+31+\left(16.4\right)=104\left(\dfrac{g}{mol}\right)\)

\(\%_H=\dfrac{3.3.1}{104}.100=8,6\%\\ \%PO_4=\dfrac{31+64}{104}.100=91,4\%\)

\(M_{10Ca\left(OH\right)_2}=10.40+\left(16+1\right).2=434\left(\dfrac{g}{mol}\right)\)

\(\%_{Ca}=\dfrac{10.40}{434}=92,2\%\\ \%_{OH}=\dfrac{\left(16+1\right).2}{434}=7,8\%\)

Answer 2

H₃PO₄ \(\left\{{}\begin{matrix}\%m_H=\dfrac{3}{98}.100\%=3,06\%\\\%m_P=\dfrac{31}{98}.100\%=31,63\%\\\%m_O=\left(100-3,06-31,63\right)\%=65,31\%\end{matrix}\right.\)

Ca(OH)₂ \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{40}{74}.100\%=54,05\%\\\%m_H=\dfrac{2}{74}.100\%=2,7\%\\\%m_O=\left(100-54,05-2,7\right)\%=43,25\%\end{matrix}\right.\)