Question

hãy tính % theo khối lượng của nguyên tố có trong phản ứng trong hợp chất:

a) NaNO\(_3\)

b) Al\(_2\)(CO\(_3\))\(_3\)

c) NH\(_4\)NO\(_3\)

Answer 1

\(a,\%Na=\dfrac{23}{85}.100\%=27,06\%\\ \%N=\dfrac{14}{85}.100\%=16,47\%\\ \%O=100\%-27,06\%-16,47\%=56,47\%\\ b,\%Al=\dfrac{54}{234}.100\%=27,1\%\\ \%C=\dfrac{36}{234}.100\%=15,4\%\\ \%O=100\%-27,1\%-15,4\%=57,5\%\)

\(c,\%N=\dfrac{28}{79}.100\%=35,4\%\\ \%H=\dfrac{4}{79}.100\%=5,1\%\\ \%O=100\%-35,4\%-5,1\%=59,5\%\)

Answer 2

\(M_{NaNO_3}=23+14+16.3=85\left(\dfrac{g}{mol}\right)\\ \Rightarrow\%m_{Na}=\dfrac{23.100\%}{85}=27\%\\ \%m_N=\dfrac{14.100\%}{85}=16,47\%\\ \Rightarrow\%m_O=100\%-\left(16,47\%+27\%\right)=56,53\)