Dolomite is a mixed carbonate of calcium and magnesium. Calcium and Magnesium carbonates both decompose upon heating to produce the metal oxides (MgO and CaO) and Carbon dioxide. If 4.84g of residue consisting of MgO and CaO remains when 9.66g of dolomite is heated until decomposition is complete, what percentage by mass of the original sample was MgCO3?
How do you setup this question? I'm completely confused...
Thanks
Chemical Principles F.24 HELP!
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 53
- Joined: Fri Sep 29, 2017 7:07 am
- Been upvoted: 1 time
-
- Posts: 36
- Joined: Sat Jul 22, 2017 3:01 am
Re: Chemical Principles F.24 HELP!
To begin you would write the two separate chemical equations, like this:
4.84 g MgO + CaO and 9.66 g MgCO3 + CaCO3
Next you just sub in x and y for the moles of each:
x mol MgCO3 and y mol CaCO3
Then plug these into two separate equations using the molar masses of each compound:
x(M(MgCO3)) + y(M(CaCO3)) = 9.66 g and x(M(MgO)) + y(M(CaO)) = 4.84 g
From here you can find the molar masses of each and plug in to solve.
Hope this helps :)
4.84 g MgO + CaO and 9.66 g MgCO3 + CaCO3
Next you just sub in x and y for the moles of each:
x mol MgCO3 and y mol CaCO3
Then plug these into two separate equations using the molar masses of each compound:
x(M(MgCO3)) + y(M(CaCO3)) = 9.66 g and x(M(MgO)) + y(M(CaO)) = 4.84 g
From here you can find the molar masses of each and plug in to solve.
Hope this helps :)
-
- Posts: 52
- Joined: Fri Sep 29, 2017 7:04 am
Re: Chemical Principles F.24 HELP!
According to the description, we get the reaction equation MgCO3 + CaCO3 → MgO + CaO + CO2. Then we balance it: MgCO3 + CaCO3 → MgO + CaO + 2CO2. According to the law of conservation of mass, the mass of products has equal to the mass of reactants, which is 9.66g. The products are Mgo, CaO, and C02. The mass of MgO and CaO is 4.84g according to question. So 9.66g - 4.84 = 4.82 is the mass of CO2. By using molar mass, we can get the number of mole of CO2, which is 0.11 mole . According to reaction equation, 2 mole of CO2 requires 1 mole of MgCO3. So we have 0.11mole / 2 =
0.055 mole of MgCO3. Then the mass of MgCO3 is 4.6g and the mass percentage is 47.6%. Hope this helps.
0.055 mole of MgCO3. Then the mass of MgCO3 is 4.6g and the mass percentage is 47.6%. Hope this helps.
Return to “Empirical & Molecular Formulas”
Who is online
Users browsing this forum: No registered users and 7 guests