Đổi 100ml = 0,1 lít
Ta có: \(n_{Na_2SO_4}=1.0,1=0,1\left(mol\right)\)
Ta lại có: \(C_{\%_{BaCl_2}}=\dfrac{m_{BaCl_2}}{100}.100\%=41,6\%\%\)
=> \(m_{BaCl_2}=41,6\left(g\right)\)
=> \(n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)
PTHH: \(Na_2SO_4+BaCl_2--->BaSO_4\downarrow+2NaCl\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\)
Vậy BaCl2 dư.
Theo PT: \(n_{BaSO_4}=n_{Na_2SO_4}=0,1\left(mol\right)\)
=> \(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)