Bài 1:
Ta có: \(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)
PT: \(2HCl+Na_2CO_3\rightarrow2NaCl+CO_2+H_2O\)
______0,1_____________0,1_____0,05 (mol)
⇒ m dd sau pư = 3,65 + 200 - 0,05.44 = 201,45 (g)
\(\Rightarrow C\%_{NaCl}=\dfrac{0,1.58,5}{201,45}.100\%\approx2,9\%\)
Bài 2:
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)
____0,1____________0,1______0,1 (mol)
⇒ m dd sau pư = 6,5 + 270 - 0,1.64 = 270,1 (g)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{0,1.161}{270,1}.100\%\approx5,96\%\)