BT8:
\(m_{Fe}=\dfrac{28\%.400}{100\%}=112g\)
\(\Rightarrow n_{Fe}=\dfrac{112}{56}=2mol\)
\(m_S=\dfrac{24\%.400}{100\%}=96g\)
\(\Rightarrow n_S=\dfrac{90}{32}=3mol\)
\(m_O=\dfrac{48\%.400}{100\%}=192g\)
\(\Rightarrow n_O=\dfrac{192}{16}=12mol\)
Vậy CTHH của Y: \(Fe_2SO_{12}\)
BT7:
\(PTK_{Q\text{ξ }tnhom}=1,275.\left(16+64\right)=102đvc\)
Mà: \(m_{Al}=\dfrac{4,5}{4}.m_{O_2}\)
Có: \(m_{Al}+m_{O_2}=102đvc\)
\(\Leftrightarrow\dfrac{4,5}{3}.m_{O_2}+m_{O_2}=102đvc\)
\(\Leftrightarrow M_{O_2}\left(\dfrac{4,5}{4}+1\right)=102đvc\)
\(\Leftrightarrow m_{O_2}=\dfrac{102}{\dfrac{4,5}{4}+1}\)
\(\Leftrightarrow m_{O_2}=48\left(1\right)\)
\(\rightarrow m_{Al}=102-48=54\left(2\right)\)
\(\left(1\right)\Rightarrow1molOxi:n_{O_2}=\dfrac{48}{16}=3mol\)
\(\left(2\right)\Rightarrow1mol\) nhôm: \(n_{Al}=\dfrac{54}{27}=2\)
Vậy CTHH là \(Al_2O_3\)
