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
No.
6.01 g/molar mass K2SO4 = moles K2SO3.
moles KCl = 2x that.
(KCl) = mols/0.99
To find the molarity (M) of KCl produced, we need to follow these steps:
1. Calculate the number of moles of K2SO3.
Given the mass of K2SO3 is 6.01 g and the molecular weight of K2SO3 is 174.27 g/mol, we can find the number of moles of K2SO3 by dividing the mass by the molecular weight:
Moles of K2SO3 = 6.01 g / 174.27 g/mol
2. Calculate the number of moles of KCl produced.
From the balanced equation, we can see that the mole ratio between K2SO3 and KCl is 1:2. Therefore, the number of moles of KCl produced is twice the number of moles of K2SO3 calculated in step 1:
Moles of KCl = Moles of K2SO3 x 2
3. Find the molarity (M) of KCl.
Molarity is defined as moles of solute divided by the volume of the solution in liters. The total volume of the reaction solution is given as 990 mL, which is equivalent to 0.99 L (since 1 L = 1000 mL). Therefore, we can calculate the molarity of KCl as follows:
Molarity (M) = Moles of KCl / Volume of solution (L)
Molarity (M) = (Moles of KCl) / 0.99 L
After following these steps and plugging in the values from the question, you can find the molarity (M) of KCl produced.