Question
If 6.01 g of K2SO3 dissolved in 930 mL of water reacts stoichiometrically according to the balanced equation in a reaction solution with a total volume of 990 mL, what is the molarity (M) of KCl produced?
K2SO3(aq) + 2 HCl(aq) → SO2(g) + 2KCl(aq) + H2O(l)
6.01gx molecular weigt of K2SO3=mole of K2SO3
mole of K2SO3x2=mole of KCL
mole of KCL/.99L=M
K2SO3(aq) + 2 HCl(aq) → SO2(g) + 2KCl(aq) + H2O(l)
6.01gx molecular weigt of K2SO3=mole of K2SO3
mole of K2SO3x2=mole of KCL
mole of KCL/.99L=M
Answers
DrBob222
No.
6.01 g/molar mass K2SO4 = moles K2SO3.
moles KCl = 2x that.
(KCl) = mols/0.99
6.01 g/molar mass K2SO4 = moles K2SO3.
moles KCl = 2x that.
(KCl) = mols/0.99