A 5.00 gram mixture of CaO and CaO3 is heated strongly. After heating, the mixture it has a mass of 4.00 grams. What % of the original mixture is CaCo3?
I assume you mean CaO and CaCO3
I am not certain what heated strongly means.
If one heats CaCO3, CaCO3>>CaO+ CO2
This is the way quicklime and cement is made.
So, if this is the reaction, you got 4 grams (molesCO2=4/44) CO2.
so if you got 1/11 moles CO2, you must have decomposed 1/11 moles CaCO3. How many grams is that?
To determine the percentage of CaCO3 in the original mixture, we need to first find the mass of CaCO3 that remains after heating.
Let's assume the mass of CaCO3 in the original mixture is x grams. Since the total mass of the mixture is 5.00 grams, the mass of CaO would be (5.00 - x) grams.
During heating, both CaO and CaCO3 will undergo a reaction, and part of the mass will be lost. Given that after heating, the mixture has a mass of 4.00 grams, we can set up the following equation:
Mass of CaO + Mass of CaCO3 = Mass after heating
(5.00 - x) grams + x grams = 4.00 grams
Now, let's solve the equation for x to find the mass of CaCO3 remaining after heating:
5.00 grams - x grams + x grams = 4.00 grams
Simplifying the equation: 5.00 grams = 4.00 grams
Subtracting 5.00 grams from both sides: x grams = 1.00 grams
So, after heating, 1.00 gram of CaCO3 remains from the original 5.00 gram mixture.
To find the percentage of CaCO3 in the original mixture, we divide the mass of CaCO3 by the total mass of the original mixture and multiply by 100:
Percentage of CaCO3 = (1.00 grams / 5.00 grams) * 100
Percentage of CaCO3 = 20%
Therefore, 20% of the original mixture is CaCO3.