A 20.29 g sample of impure MgCO3 is completely decomposed at 1000oC in previously evacuated 2.0 L reaction vessel. After the reaction was complete the solid residue had a mass of 15.9 g. Assume that only MgCO3 could produce gas CO2 what percent of original sample was magnesium carbonate? What was the pressure of the CO2 produced?
Ok, so everyone I have talked to has different ideas on how I should answer this question.
For part (A) the ideas have been.
(1) 15.9/20.29= 78.36%
(2) (20.29 - 15.9)/20.29 = 21.64%
(3) Use g of CO2 to find g of MgNO3. 4.39 g CO2 * (1 mol/44g) * (1 mol/1 mol) * (84g / 1 mol) = 8.38g
Then, 8.38/20.29 = 41.3%
Do any of those seem right? I also have ideas for part b, but I will write them later.
4 answers
It looks OK. For the 2nd question convert the grams of CO2 to moles and use the Ideal Gas Law to find the pressure due to CO2.
I gave three different answers. Which one looks OK?
Sorry. The 3rd one looks OK.
That is what I am thinking. Thanks.
1 answer