a) \(n_{CuSO_4}=\dfrac{20}{160}=0,125\left(mol\right)\)
Ta có: \(n_{Cu}=n_S=n_{CuSO_4}=0,125\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,125\times64=8\left(g\right)\)
\(\Rightarrow\%m_{Cu}=\dfrac{8}{20}\times100\%=40\%\)
\(m_S=0,125\times32=4\left(g\right)\)
\(\Rightarrow\%m_S=\dfrac{4}{20}\times100\%=20\%\)
Ta có: \(n_O=4n_{CuSO_4}=4\times0,125=0,5\left(mol\right)\)
\(\Rightarrow m_O=0,5\times16=8\left(g\right)\)
\(\Rightarrow\%m_O=\dfrac{8}{20}\times100\%=40\%\)
b) \(m_{MgCO_3}=2\times84=168\left(g\right)\)
Ta có: \(n_{Mg}=n_C=2\left(mol\right)\)
\(\Rightarrow m_{Mg}=2\times24=48\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\dfrac{48}{168}\times100\%=28,57\%\)
\(m_C=2\times12=24\left(g\right)\)
\(\Rightarrow\%m_C=\dfrac{24}{168}\times100\%=14,29\%\)
Ta có: \(n_O=3n_{MgCO_3}=3\times2=6\left(mol\right)\)
\(\Rightarrow m_O=6\times16=96\left(g\right)\)
\(\Rightarrow\%m_O=\dfrac{96}{168}\times100\%=57,14\%\)
c) \(n_{N_2O_5}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow m_{N_2O_5}=0,3\times108=32,4\left(g\right)\)
Ta có: \(n_N=2n_{N_2O_5}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m_N=0,6\times14=8,4\left(g\right)\)
\(\Rightarrow\%m_N=\dfrac{8,4}{32,4}\times100\%=25,93\%\)
Ta có: \(n_O=5n_{N_2O_5}=5\times0,3=1,5\left(mol\right)\)
\(\Rightarrow m_O=1,5\times16=24\left(g\right)\)
\(\Rightarrow\%m_O=\dfrac{24}{32,4}\times100\%=74,07\%\)
