Bài 1:
Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
____0,15____0,3_____0,15___0,15 (mol)
\(V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
\(C\%_{HCl}=\dfrac{0,3.36,5}{400}.100\%=2,7375\%\)
Ta có: m dd sau pư = 8,4 + 400 - 0,15.2 = 408,1 (g)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{0,15.127}{408,1}.100\%\approx4,7\%\)
Bài 2:
\(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,2_____0,4_____0,2___0,2 (mol)
\(C\%_{HCl}=\dfrac{0,4.36,5}{500}.100\%=2,92\%\)
Ta có: m dd sau pư = 0,2.65 + 500 - 0,2.2 = 512,6 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,2.136}{512,6}.100\%\approx5,3\%\)
Bài 3:
Ta có: \(n_{NaOH}=\dfrac{200.10\%}{40}=0,5\left(mol\right)\)
PT: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
_______0,5_______0,25______0,25_______0,25 (mol)
\(a=C\%_{CuSO_4}=\dfrac{0,25.160}{320}.100\%=12,5\%\)
Ta có: m dd sau pư = 200 + 320 - 0,25.98 = 495,5 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,25.142}{495,5}.100\%\approx7,2\%\)