A 1.00 g sample of enriched water, a mixture of H2O and D2O, reacted completely with Cl2 to give a mixture of HCl and DCl. The HCl and DCl were then dissolved in pure H2O to make a 1.00 L solution. A 25.00 mL sample of the 1.00 L solution was reacted with excess AgNO3 and 0.3800 g of an AgCl precipitate formed. What was the mass % of D2O in the original sample of enriched water?

Question

A 1.00 g sample of enriched water, a mixture of H2O and D2O, reacted completely with Cl2 to give a mixture of HCl and DCl. The HCl and DCl were then dissolved in pure H2O to make a 1.00 L solution. A 25.00 mL sample of the 1.00 L solution was reacted with excess AgNO3 and 0.3800 g of an AgCl precipitate formed. What was the mass % of D2O in the original sample of enriched water?

I am confused with how to go about answering this question, if someone could help me it would be greatly appreciated.

Answer 1

I have deleted the original response and replaced it with the following. Check my thinking.

Cl2 + H2O ==> HOCl + HCl
Cl2 + D2O ==> DOCl + DCl

let x = mass H2O
and y = mass D2O
------------------
x + y = 1.00
[x*(molar mass AgCl/molar mass H2O)] + [y*(molar mass AgCl/molar mass D2O)] = 0.3800 x 1000/25

Two equation in two unknowns. Solve for x and y and convert to percent. Check my thinking.

Answer 2

To find the mass % of D₂O in the original sample of enriched water, we need to understand the steps involved in the reaction and use stoichiometry and mass balance.

Let's break down the steps:

Step 1: Reaction of enriched water with Cl₂
The balanced chemical equation for this reaction is:
H₂O + Cl₂ -> 2HCl

Given that the initial mass of the sample is 1.00 g, and the products are HCl and DCl, we can infer that the mass of HCl and DCl produced would be equal to 1.00 g.

Step 2: Dissolution of HCl and DCl in H₂O
In this step, the HCl and DCl combine with pure H₂O to form a 1.00 L solution.

Step 3: Reaction of the 1.00 L solution with AgNO₃
The balanced chemical equation for this reaction is:
AgNO₃(aq) + HCl(aq) -> AgCl(s) + HNO₃(aq)

Since a 25.00 mL sample of the 1.00 L solution was reacted and 0.3800 g of AgCl precipitate was formed, we can use stoichiometry to find the moles of HCl reacted.

Step 4: Calculate the moles and mass of D₂O
Now, we need to determine how many moles of D₂O were present in the original enriched water sample. To do this, we'll use the mole ratio of HCl to D₂O in the balanced equation from step 1.

Step 5: Calculate mass % of D₂O
Finally, to determine the mass % of D₂O in the original sample, we'll divide the mass of D₂O by the total mass of the original sample and multiply by 100.

By following these steps, we can find the mass % of D₂O in the original sample of enriched water.