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.
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.
To find the percent of the original sample that is magnesium carbonate (MgCO3) in this question, you need to compare the mass of MgCO3 to the total mass of the sample.
Here's how you can calculate it:
1. Calculate the mass of MgCO3 in the sample:
Mass of MgCO3 = Mass of sample - Mass of residue
Mass of MgCO3 = 20.29 g - 15.9 g = 4.39 g
2. Calculate the percent of MgCO3 in the original sample:
Percent of MgCO3 = (mass of MgCO3 / mass of sample) x 100%
Percent of MgCO3 = (4.39 g / 20.29 g) x 100% ≈ 21.64%
Therefore, the correct answer for part (A) is option (2): Approximately 21.64% of the original sample is magnesium carbonate (MgCO3).
Now, let's move on to part (B) to calculate the pressure of the CO2 produced.
To find the pressure of the CO2 gas produced, we can use the ideal gas law equation, which is PV = nRT. However, since we don't have the values of pressure (P), volume (V), or temperature (T) given in the question, we'll use a different approach.
1. Convert the mass of CO2 produced to moles:
Moles of CO2 = Mass of CO2 / molar mass of CO2
Molar mass of CO2 = 12.01 g/mol (carbon) + 2*(16.00 g/mol) (oxygen) = 44.01 g/mol
Moles of CO2 = 4.39 g / 44.01 g/mol ≈ 0.10 mol
2. Calculate the number of moles of CO2 produced using the ideal gas law:
n = PV / RT
Here, we have R = 0.0821 L·atm/(mol·K), and the temperature is 1000°C, which needs to be converted to Kelvin (K) by adding 273.15.
Let's assume the pressure is P.
0.10 mol = P * 2.0 L / (0.0821 L·atm/(mol·K) * (1000 + 273.15) K)
Solving this equation for P:
P ≈ 10.17 atm
Therefore, the pressure of the CO2 produced is approximately 10.17 atm.
I hope this explanation helps clarify the correct calculations for both part (A) and part (B) of the question.