Question

Tính thành phần trăm các nguyên tố có trong Na₂CO₃

Answer 1

Ta có: \(M_{Na_2CO_3}=23.2+12+3.16=106\left(\dfrac{g}{mol}\right)\)

\(\%m_{Na}=\dfrac{2.23}{106}.100=43,396\%\\ \%m_C=\dfrac{12}{106}.100=11,321\%\\ \%m_O=100\%-43,396\%-11,321\%=45,283\%\)

Answer 2

Ta có:\(M_{Na_2CO_3}=106\)(g/mol)

\(\Rightarrow\%M_{Na}=\frac{46}{106}.100\%=43,4\left(\%\right)\)

\(\Rightarrow\%M_C=\frac{12}{106}.100\%=11,3\left(\%\right)\)

\(\Rightarrow\%M_O=100\%-43,4\%-11,3\%=45,3\left(\%\right)\)