We know that one mole of any gas at STP occupies 22.4 liters of volume. We also know that one mole of carbon dioxide is 44.01 grams of CO2. If there are 44.01 grams of this gas in 22.4 liters at STP, then there will be about 0.98 grams of CO2 in half a liter (500 ml) of the gas at STP.
22 grams carbon dioxide (0.5 moles)
33
The mass of carbon dioxide is 141,2 g.
If the two occupy the same volume, 25 grams is denser.
If all of the quantities stated actually reacted, the law of the conservation of mass shows that the mass of carbon dioxide produced would be 40 - 18 or 22 grams.
Weight:1.45 grams volume: 6.789
Moles of carbon dioxide = grams/amu of carbon dioxide. Moles = 19g/44amu Moles of carbon dioxide = .432
22 grams carbon dioxide (0.5 moles)
33
6 grams
The mass of carbon dioxide is 141,2 g.
The volume of carbon dioxide is 8,4 L at oC.
If the two occupy the same volume, 25 grams is denser.
To determine the volume of carbon dioxide needed, you would need to know the stoichiometry of the reaction between carbon dioxide and calcium carbonate. In this case, since 20 grams of calcium carbonate is given, you would convert that to moles using the molar mass of calcium carbonate. Then, using the balanced equation, you can determine the mole ratio between carbon dioxide and calcium carbonate. Finally, using the molar volume of carbon dioxide gas at the given conditions (usually 22.4 L/mol at standard temperature and pressure), you can calculate the volume of carbon dioxide needed.
11 grams because all is reacted and there is no reactant left over, although if there were only 3 grams of carbon there would have to be 6 grams of oxygen for this to be viable as carbon dioxide is CO2 so the question asked was itself wrong.
Look up the molecular weight of carbon dioxide in the periodic table. The formula for carbon dioxide is CO2, which means one atom of carbon and two atoms or oxygen per molecule of carbon dioxide. Carbon has molecular weight of 12. Oxygen molecular weight is 16. Total 12+16+16= 44 11 grams/44 grams/mole=0.25 moles of carbon The grams of water and combustion of 7.5 grams are totally irrelevant. They are only given to possibly confuse you.
no