Magnesium oxide is not very soluble in water, and is is difficult to titrate directly. It's purity can be determined by use of a back-titration method. 4.06 g of impure magnesium oxide was completely dissolved in an excess of 0.20 mol/L for neutralization. Calculate the mass of magnesium oxide that reacted with the initial hydrochloric acid, hence the % purity of magnesium oxide.
Please help with a good procedure. Thanks.
You've left out part of the problem. How much of the 0.20 M acid( what acid or at least is it monoprotic or diprotic). Then you don't give any numbers for the back titration. Look over your post and fill in the blanks.
To determine the mass of magnesium oxide that reacted with the initial hydrochloric acid and calculate the % purity, we will use the back-titration method. Here's how we can solve this problem step by step:
1. Write the balanced chemical equation for the reaction:
MgO + 2HCl -> MgCl2 + H2O
2. Calculate the number of moles of hydrochloric acid used:
Moles of HCl = Concentration of HCl (mol/L) * Volume of HCl (L)
As an excess of HCl was used, the number of moles of HCl used is equal to the initial volume of HCl used.
3. Calculate the number of moles of magnesium oxide reacted with the HCl:
From the balanced equation, we can see that 1 mole of MgO reacts with 2 moles of HCl.
So, Moles of MgO = 0.5 * Moles of HCl
4. Calculate the mass of magnesium oxide reacted:
Mass of MgO = Moles of MgO * Molar mass of MgO
5. Calculate the % purity of magnesium oxide:
% Purity = (Mass of pure MgO / Mass of impure MgO) * 100
Now let's substitute the given values into the formulas:
Given:
- Mass of impure MgO = 4.06 g
- Concentration of HCl = 0.20 mol/L
Step 2:
Moles of HCl = 0.20 mol/L * Volume of HCl
Since the volume of HCl is not given, we cannot calculate the exact moles of MgO.
Therefore, without the volume of HCl used, we cannot determine the moles of MgO that reacted or calculate the % purity of magnesium oxide.