What is the mass of each compound in the original mixture when heated to 700C?

  • Thread starter Thread starter leolaw
  • Start date Start date
  • Tags Tags
    Gas Molecule
Click For Summary
SUMMARY

The discussion focuses on determining the mass of calcium chlorate (Ca(ClO3)2) and calcium hypochlorite (Ca(ClO)2) in a 10.00 gram mixture after heating to 700°C, resulting in the decomposition of both compounds and the release of oxygen gas. Using the ideal gas law, the total oxygen collected was calculated to be 0.1252 mol. The balanced equations for the decomposition reactions were established, leading to the conclusion that approximately 0.0313 mol of each compound contributes to the total mass, confirming the original mixture's mass is consistent with the calculated values.

PREREQUISITES
  • Understanding of stoichiometry and balanced chemical equations
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of molar mass calculations
  • Basic principles of chemical decomposition reactions
NEXT STEPS
  • Study the ideal gas law applications in chemical reactions
  • Learn about stoichiometric calculations in chemical mixtures
  • Explore the properties and reactions of calcium compounds
  • Investigate methods for verifying chemical reaction yields
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical analysis, particularly those focusing on stoichiometry and gas laws in reaction scenarios.

leolaw
Messages
85
Reaction score
1
I am stuck in this gas problem:
You are given a 10.00 gram solid mixture of Ca(ClO_3)_2 and Ca(ClO)_2. When this mixture is heated, both componds decomplse, releasing oxygen gas and leaving behind solid CaCl_2

When this 10.00gram mixture is sealed in a 10.0-Liter evacuated vessel and heated to 700C, both compounds completely decompose according to the above reactions. The final pressure in the vessel is 1.00atm. Determine the mass of each compound in the original mixture.

Ok. I have written out the balanced equation of the two substances:
Ca(ClO_3)_2 + O_2 --> CaCl_2 + 4O_2 and
Ca(ClO)_2 + O_2 --> CaCl_2 + 2O_2

using the ideal gas law, i have found that the total O_2 collected is 0.1252 mol (V = 10L , T = 973K, P = 1.00atm)

so I have let x equal to the gram of Ca(ClO_3)_2 and y equal to the gram of Ca(ClO)_2, combine them toegether, I have:
x + y = 10

and then i found out the precentage of O2 that each substance produce, so I have another equation:
(0.0193) x + (0.0186) y = 0.1252

when i solve for x and y , i have x equal to -86 and y = 96, which obviously, is not possible?

Any clues on how to solve this probelm?
 
Chemistry news on Phys.org
From the way the problem is worded I think you equations should be

Ca(ClO_3)_2 --> CaCl_2 + 3O_2
Ca(ClO)_2 --> CaCl_2 + O_2

Lets assume you have 1 mol of each starting material. Then according to the balanced equations, after decomposition Ca(ClO3)2 will give off 3 moles of O2 and Ca(ClO)2 will give off 1 mole of O2 for a total of 4 moles of oxygen. In other words 3/4 of the total number of moles come from Ca(ClO3)2 and 1/4 of the total number of moles of O2 come from Ca(ClO)2. You found .1252mol of total oxygen, so 3/4th of that would be .0939 moles of O2 coming from Ca(ClO3)2 and 1/4ths of that would be .0313 moles of O2 that would be coming from Ca(ClO)2.


Using the balanced equations again-

0.0939 mol O2 x 1molCa(ClO3)2/3mol O2=.0313 mol Ca(ClO3)2
and
.0313 mol O2 x 1molCa(ClO)2/1mol O2=0.0313 mol Ca(ClO)2

Now to double check that you are right you can add up the moles of each substance in grams and you should get 10.0g.

mw of Ca(ClO3)2 206.98g/mol
and mw of Ca(ClO)2 is 142.98 g/mol

thus .0313 mol(206.98g/mol)+.0313mol(142.98g/mol)=10.0g (approximately due to rounding). So you know you have to be right
 
Yes, I have fixed up on the equation.
THx
 

Similar threads

Solving Gas Mixture Problem: 10g Ca(ClO3)2 + Ca(ClO)2
  • · Replies 3 ·
Replies
3
Views
8K