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D = mdd/V ---> mdd = D.V = 1,28.200 = 256 gam. ---> mCaCl2 = mdd.C%/100 = 256.30/100 = 76,8gam.
`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`
0,0625----------0,1875---------0,0625 mol
`->n_(Fe_2O_3)=10/160=0,0625mol`
`->m_(Fe_2(SO_4)_3)=0,0625.400=25g`
`->C%(H_2SO_4)=((0,1875.98)/(450)).100%=4,083%`
`#YBtran<3`
\(n_{Fe_2O_3}=\dfrac{10}{160}=0,0625\left(mol\right)\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,0625\left(mol\right)\\ a,m=m_{Fe_2\left(SO_4\right)_3}=400.0,0625=25\left(g\right)\\ b,n_{H_2SO_4}=3.0,0625=0,1875\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,1875.98}{450}.100\%\approx4,083\%\)
Ta có: \(n_{H_2SO_4}=\dfrac{44,1}{98}=0,45\left(mol\right)\)
\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
PT: \(H_2SO_4+CaCl_2\rightarrow2HCl+CaSO_{4\downarrow}\)
\(H_2SO_{4\left(dư\right)}+2NaOH\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(pư\right)}=0,45-0,25=0,2\left(mol\right)\)
Theo PT: \(n_{CaSO_4}=n_{H_2SO_4\left(pư\right)}=0,2\left(mol\right)\)
\(\Rightarrow m_{CaSO_4}=0,2.136=27,2\left(g\right)\)
Bảo toàn KL: \(m_{Na_2SO_4}+m_{CaCl_2}=m_{NaCl}+m_{CaSO_4}\)
\(\Rightarrow m_{CaSO_4}=14,2+11,1-17=8,3\left(g\right)\)
\(m_{ct}=\dfrac{m_{dd}\times C}{100}=\dfrac{500\times10}{100}=50\\ \Rightarrow A\)