A 21.5997 g sample of impure magnesium carbonate was heated to complete decomposition
according to the equation
MgCO3(s) → MgO(s) + CO2(g).
After the reaction was complete, the solid
residue (consisting of MgO and the original
impurities) had a mass of 15.5207 g. Assuming that only the magnesium carbonate had
decomposed, how much magnesium carbonate was present in the original sample?
Answer in units of g
21.5997-15.207 = ? = mass CO2
g CO2 x (1 mol MgCO3/1 mol CO2) = g MgCO3.
I don't know eather!
That should be 21.5997-15.5207 = mass CO2.
I assume you mean either.
To find out how much magnesium carbonate was present in the original sample, we need to calculate the mass of magnesium carbonate that decomposed during the reaction.
First, let's calculate the mass of magnesium carbonate that remained after the reaction:
Mass of residue = 15.5207 g
Since the residue consists of MgO (magnesium oxide) and impurities, we need to subtract the mass of MgO from the residue to find the mass of the impurities.
To do this, we need to know the atomic masses of magnesium (Mg), carbon (C), and oxygen (O) from the periodic table:
Atomic mass of Mg = 24.305 g/mol
Atomic mass of C = 12.011 g/mol
Atomic mass of O = 15.999 g/mol
Next, let's calculate the moles of MgO present in the residue:
Molar mass of MgO = Atomic mass of Mg + Atomic mass of O
Molar mass of MgO = 24.305 g/mol + 15.999 g/mol
Molar mass of MgO = 40.304 g/mol
Moles of MgO = Mass of MgO / Molar mass of MgO
Moles of MgO = 15.5207 g / 40.304 g/mol
Now, let's calculate the moles of MgO.
Molar ratio of MgCO3 to MgO can be obtained from the balanced chemical equation:
1 mol of MgCO3 reacts to form 1 mol of MgO
Therefore, the moles of MgCO3 that decomposed are equal to the moles of MgO produced.
Finally, let's convert the moles of MgCO3 to grams:
Molar mass of MgCO3 = Atomic mass of Mg + Atomic mass of C + 3 * Atomic mass of O
Molar mass of MgCO3 = 24.305 g/mol + 12.011 g/mol + 3 * 15.999 g/mol
Molar mass of MgCO3 = 84.313 g/mol
Mass of MgCO3 = Moles of MgO * Molar mass of MgCO3
Mass of MgCO3 = (15.5207 g / 40.304 g/mol) * 84.313 g/mol
Therefore, the mass of magnesium carbonate present in the original sample is:
Mass of MgCO3 = 21.5997 g - Mass of MgCO3 decomposed
Mass of MgCO3 = 21.5997 g - [(15.5207 g / 40.304 g/mol) * 84.313 g/mol]
By evaluating the above expression, the amount of magnesium carbonate present in the original sample is obtained, which gives the answer in grams.