\(M_{NaNO_3}=23+14+16\times3=85\left(g\right)\)
\(\%Na=\dfrac{23}{85}\times100\%=27,06\%\)
\(\%N=\dfrac{14}{85}\times100\%=16,47\%\)
\(\%O=100\%-27,06\%-16,47\%=56,47\%\)
\(M_{K_2CO_3}=39\times2+12+16\times3=138\left(g\right)\)
\(\%K=\dfrac{39\times2}{138}\times100\%=56,52\%\)
\(\%C=\dfrac{12}{138}\times100\%=8,7\%\)
\(\%O=100\%-8,7\%-56,52\%=34,78\%\)
\(M_{Al\left(OH\right)_3}=27+3\times17=78\left(g\right)\)
\(\%Al=\dfrac{27}{78}\times100\%=34,62\%\)
\(\%O=\dfrac{16\times3}{78}\times100\%=61,54\%\)
\(\%H=100\%-34,62\%-61,54\%=3,84\%\)
* \(M_{Na\left(NO_3\right)}=23+14+48=85\left(g/mol\right)\)
\(\%Na=\dfrac{23}{85}.100\%=27,05\%\)
\(\%N=\dfrac{14}{85}.100\%=16,47\%\)
\(\Rightarrow\%O=100\%-\left(27,05+16,47\right)\%=56,48\%\)
* \(M_{K_2CO_3}=78+12+48=138\left(g/mol\right)\)
\(\%K=\dfrac{78}{138}.100\%=56,52\%\)
\(\%C=\dfrac{12}{138}.100\%=8,69\%\)
\(\Rightarrow\%O=100\%-\left(56,52+8,69\right)\%=34,79\%\)
* \(M_{Al\left(OH\right)_3}=27+48+3=78\left(g/mol\right)\)
\(\%Al=\dfrac{27}{78}.100\%=34,61\%\)
\(\%O=\dfrac{48}{78}.100\%=61,53\%\)
\(\Rightarrow\%H=100\%-\left(34,61+61,53\right)\%=3,86\%\)
\(M_{NaNO_3}=23+14+16.3=85\left(g/mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Na=\dfrac{23.1}{85}.100=27,06\%\\\%N=\dfrac{14.1}{85}.100=16,47\%\\\%O=\dfrac{16.3}{85}.100=56,48\%\end{matrix}\right.\)
\(M_{K_2CO_3}=39.2+12+16.3=138\left(g/mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%K=\dfrac{39.2}{138}.100=56,52\%\\\%C=\dfrac{12.1}{138}.100=8,7\%\\\%O=\dfrac{16.3}{138}.100=34,78\%\end{matrix}\right.\)
\(M_{Al\left(OH\right)_3}=27+3.\left(16+1\right)=78\left(g/mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%Al=\dfrac{27.1}{78}.100=34,61\%\\\%O=\dfrac{16.3}{78}.100=61,53\%\\\%H=\dfrac{1.3}{78}.100=3,86\%\end{matrix}\right.\)
