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Author Topic: Lab questions - Stoichiometry  (Read 853 times)

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itslindahuynh

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Lab questions - Stoichiometry
« on: December 04, 2012, 12:44:47 PM »

I have two questions that I am completely clueless on.

The first:
Problem: What is the mass of sodium silicate in a 25.0 mL sample of the solution used in a chemical process?
Prediction: If the process is operating as expected, the mass of sodium silicate in a 25.0 mL sample should always be between 6.40 g and g.
Design: An excess quantity of iron (III) nitrate is added to the sample of sodium silicate. The resulting precipitate is separated by filtration. After the precipitate has dries, the mass is determined.
Evidence:
mass of filter paper: .98 g
mass of dried filter paper plus precipitate: 9.45 g
The color of the filtrate was yellow-orange.

For that question, I wrote out the balanced chemical equation and wrote that Iron (III) Nitrate has a mass of 8.46g and that it was a product. Sodium silicate is a reactant and 25.0 mL was used. After that, I have no clue how to proceed.

The second question:
Problem: What is the amount concentration of silver nitrate in the solution to be recycled?
Evidence: A white precipitate was formed in the reaction with acqueous sodium sulfate.
voluume of silver nitrate solution: 100 mL
mass of filter paper: 1.27 g
mass of dried filter paper plus precipitate: 6.74g

For this question, I have no idea what the products or reactants are...

Thanks for your help.

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Hunter2

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Re: Lab questions - Stoichiometry
« Reply #1 on: December 04, 2012, 07:55:01 PM »

For the first question. The precipitate is not iron-III-nitrate it is iron-III-silicate. If you know the mass then you can calculate the moles. Check in your equation how much sodium silicate correspond to it. Backwards you can calculate mass of it. The you have to get the concentration if you calculate with your volume.

Second question: If you have silver ion in a solution and you put sulfate to it, what happen? More better it would work with chloride.
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Dan

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Re: Lab questions - Stoichiometry
« Reply #2 on: December 04, 2012, 09:05:10 PM »

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