The answer is 19,288 g cacium carbonate.
1 mole of sodium carbonate + 2 moles of Hydrochloric acid would produce 1 mole of Carbon Dioxide which would occupy 22.4 liters at standard temperature and pressure
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
The volume of carbon dioxide is 91,9 L at 0 oC.
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.
It doesn't exist - calcium carbonate is limestone/marble - its insoluble
1 mole of sodium carbonate + 2 moles of Hydrochloric acid would produce 1 mole of Carbon Dioxide which would occupy 22.4 liters at standard temperature and pressure
If 15 liters of propane are completely consumed 90,25 grams of carbon dioxide are produced.
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
88
The volume of CO2 is 4,94 L.
The volume of carbon dioxide is 91,9 L at 0 oC.
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.
contains the same number of molecules
Average coffin size is 84 inches; width 28 inches, and height 23 inches which is 2.08m 0.71m, 0.58m which is 856 liters A part of it is filled (with you and other stuff) so lets lose 30%. 856*0.7 = 599 liters Your problem will be carbon dioxide buildup. You can cycle air through once before carbon dioxide levels go above 5% and you have real problems. At this point oxygen level is around 16% which is enough, though not comfortable. But carbon dioxide is making damage in your body and you will be probably dead when carbon dioxide level exceeds 7%. Quite painfully I might add. Human consumes 450 liters of air per hour. So you are in serious trouble in 1.3 hours and dead before 2 hours. At-least I think so, though I am not an expert.
To determine the number of liters of carbon dioxide produced in this reaction, we need the balanced equation and the molar mass of carbon dioxide. The balanced equation is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O The molar mass of CO2 is 44.01 g/mol. First, we calculate the number of moles of CaCO3: 906 g / molar mass of CaCO3 = moles of CaCO3 Using the balanced equation, we see that the stoichiometric coefficient of CO2 is 1. This means that the number of moles of CO2 produced is equal to the number of moles of CaCO3. Finally, we convert moles of CO2 to liters using the ideal gas law: moles of CO2 x 22.4 L/mol = liters of CO2. Therefore, the number of liters of CO2 produced from 906 grams of CaCO3 can be calculated as follows: liters of CO2 = (906 g / molar mass of CaCO3) x 22.4 L/mol
124,9 g grams of ammonium carbonate are needed.