a) \(M_{Ca\left(OH\right)_2}=40+2\times\left(16+1\right)=74\left(g\right)\)
\(\Rightarrow\%Ca=\dfrac{40}{74}\times100\%=54,05\%\)
\(\%O=\dfrac{16\times2}{74}\times100\%=43,24\%\)
\(\Rightarrow\%H=100\%-54,05\%-43,24\%=2,71\%\)
b) 1) \(n_{HNO_3}=\dfrac{63}{63}=1\left(mol\right)\)
Ta có: \(n_O=3n_{HNO_3}=3\times1=3\left(mol\right)\)
\(\Rightarrow m_O=3\times16=48\left(g\right)\)
2) \(n_{H_2O}=\dfrac{27}{18}=1,5\left(mol\right)\)
Ta có: \(n_O=n_{H_2O}=1,5\left(mol\right)\)
\(\Rightarrow m_O=1,5\times16=24\left(g\right)\)
3) \(n_{HgO}=\dfrac{2170}{217}=10\left(mol\right)\)
Ta có: \(n_O=n_{HgO}=10\left(mol\right)\)
\(\Rightarrow m_O=10\times16=160\left(g\right)\)
