Question

Tính thành phần % về khối lượng của các nguyên tố có trong hợp chất sau:

CuO, P2O5, H2SO4, Al2(SO4)3, NH4NO3, Ca3(PO4)2

Answer 1

*CuO

\(\%M_{Cu}=\frac{64}{80}.100\%=80\%\)

\(\%M_O=100-80=20\%\)

*P2O5

\(\%M_P=\frac{31.2}{142}.100\%=43,66\%\)

\(\%M_O=100-43,66=56,34\%\)

*\(H2SO4\)

\(\%M_H=\frac{2}{98}.100\%=2,04\%\)

\(\%M_S=\frac{32}{98}.100\%=32,65\%\)

\(\%M_O=100-32,65-2,04=65,31\%\)

* Al2(SO4)3

\(\%M_{Al}=\frac{27.2}{400}.100\%=13,5\%\)

\(\%M_S=\frac{32.3}{400}.100\%=24\%\)

\(\%M_O=100-24-13,5=62,5\%\)

\(\cdot NH4NO3\)

\(\%M_N=\frac{14.2}{80},100\%=35\%\)

\(\%M_H=\frac{4}{80}.100\%=5\%\)

\(\%M_O=100-35-5=60\%\)

*\(Ca3\left(PO4\right)3\)

\(\%M_{Ca}=\frac{40.3}{405},100\%=29,85\%\)

\(\%M_P=\frac{31.3}{405}.100\%=22,96\%\)

\(\%M_O=100-22,96-29,85=47,19\%\)