Question

A mixture consisting of only iron(III) chloride (FeCl3) and aluminum chloride (AlCl3) weighs 1.0397 g. When...

A mixture consisting of only iron(III) chloride (FeCl3) and aluminum chloride (AlCl3) weighs 1.0397 g. When the mixture is dissolved in water and an excess of silver nitrate is added, all the chloride ions associated with the original mixture are precipitated as insoluble silver chloride (AgCl). The mass of the silver chloride is found to be 3.0178 g. Calculate the mass percentages of iron(III) chloride and aluminum chloride in the original mixture.

Mass percent FeCl3 = %
Mass percent AlCl3 = %
0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
A mixture consisting of only iron(III) chloride (FeCl3) and aluminum chloride (AlCl3) weighs 1.0397 g. When...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A student obtained a vial of powder consisting of a mixture of iron (III) chloride hexahydrate...

    A student obtained a vial of powder consisting of a mixture of iron (III) chloride hexahydrate (FeCl_3-6H_2O) and silver nitrate (AgNO_3) in unknown proportions. He weighed the vial with the powder in it and found the mass to be 20.560 g. He then poured the powder into a beaker and found the mass of the empty vial to be 15.723 g. Water was added (500 mL) to powder in the beaker, and the contents were mixed and heated for 15...

  • Silver nitrate, AgNOs, reacts with iron(III) chloride, FeCl3,to give silver chloride, AgCl, and iron(III) nitrate, Fe(NO3)....

    Silver nitrate, AgNOs, reacts with iron(III) chloride, FeCl3,to give silver chloride, AgCl, and iron(III) nitrate, Fe(NO3). A solution containing 18.0 g of AgNO, was mixed with a solution containing 32.4 g of FeCls. How many grams of which reactant remains after the reaction is over?

  • A 18.88-g sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in...

    A 18.88-g sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form 36.80 g of solid. Determine the mass percent of sodium chloride in the original mixture.

  • A 0.5711 g sample of a pure soluble chloride compound is dissolved in water, and all...

    A 0.5711 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCI by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.2913 g. What is the mass percentage of chlorine in the original compound? 9 more group attempts remaining Submit Answer Retry Entire Group

  • A sample of a substance with the empirical formula XC1_2 weighs 0.5785 g. When it is...

    A sample of a substance with the empirical formula XC1_2 weighs 0.5785 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 1.2794 g. The chemical reaction is XCI_2 + 2 AgNO_3 rightarrow 2 AgCI + X(NO_3)_2 Calculate the formula mass of XCI_2. Formula mass XCI_2 = g mol^-1 Calculate the atomic mass of X. Atomic...

  • A sample of a substance with the empirical formula XCl3 weighs 0.4393 g. When it is...

    A sample of a substance with the empirical formula XCl3 weighs 0.4393 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 1.1926 g. The chemical reaction is XCl3 + 3 AgNO33AgCl + X(NO3)3 (a) Calculate the formula mass of XCl3. Formula mass XCl3 = g mol-1 (b) Calculate the atomic mass of X. Atomic mass...

  • A sample of a substance with the empirical formula XCl3 weighs 0.5183 g. When it is...

    A sample of a substance with the empirical formula XCl3 weighs 0.5183 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCI by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to he 1.4071 g. The chemical reaction is XCl3 +3 AgNO3 3AgCI + X(NO3)3 (a) Calculate the formula mass of XCl3. Formula mass XCI3- g mol (b) Calculate the atomic mass of X Atomic mass X-...

  • A 12.42 g sample of a mixture of NaClø and CaCl2ø) was dissolved in water. Excess...

    A 12.42 g sample of a mixture of NaClø and CaCl2ø) was dissolved in water. Excess AgNO was added and all of the chloride was precipitated as AgCl. 31.70 g of AgCI was collected. Calculate the %w/w of NaCl in the original mixture.

  • 1. A 0.4187 g sample of a pure soluble chloride compound is dissolved in water, and...

    1. A 0.4187 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 0.9536 g. What is the mass percentage of chlorine in the original compound? ________% 2. A student determines the calcium content of a solution by first precipitating it as calcium hydroxide, and then decomposing the hydroxide...

  • Experiment: Sodium Content in butter Intro: An experiment conducted to determine that amount of sodium in...

    Experiment: Sodium Content in butter Intro: An experiment conducted to determine that amount of sodium in butter through back titration of measuring chloride ions. the chloride ions themselves are not directly titrated ; rather a known excess of silver nitrate is added to the sample. A silver chloride precipitate immediately forms leaving behind the excess silver ions. the leftover silver ions are then titrated with ammonium thiocyanate forming a silver thiocyanate precipitate. Based on the amount of ammonium thiocyanate consumed,...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT