Question

Answer 1

1. \(n_{Fe\left(NO_3\right)_2}=0,35.2=0,7\left(mol\right)\Rightarrow m_{Fe\left(NO_3\right)_2}=0,7.242=169,4\left(g\right)\)

2. \(n_{Ba\left(OH\right)_2}=0,08.1,5=0,12\left(mol\right)\Rightarrow m_{Ba\left(OH\right)_2}=0,12.171=20,52\left(g\right)\)

\(n_{KOH}=0,08.1,25=0,1\left(mol\right)\Rightarrow m_{KOH}=0,1.56=5,6\left(g\right)\)

3. \(m_{K_2SO_4}=250.10,44\%=26,1\left(g\right)\Rightarrow n_{K_2SO_4}=\dfrac{26,1}{174}=0,15\left(mol\right)\)

4. \(m_{CuCl_2}=600.9\%=54\left(g\right)\Rightarrow n_{CuCl_2}=\dfrac{54}{135}=0,4\left(mol\right)\)

\(m_{MgSO_4}=600.6\%=36\left(g\right)\Rightarrow m_{MgSO_4}=\dfrac{36}{120}=0,3\left(mol\right)\)