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How many grams of CaCO3 form if 155g of water reacts?
To determine the amount of CaCO3 formed, we need to know the molar ratio of water to CaCO3 in the chemical reaction. Without this information, it is not possible to calculate the amount of CaCO3 formed when 155g of water reacts.
How many grams of calcium carbonate are needed to produce 45.0L of carbon dioxide at STP?
The balanced chemical equation for the reaction of calcium carbonate (CaCO3) to produce carbon dioxide (CO2) is: CaCO3 -> CaO + CO2. Using the ideal gas law, we can calculate the moles of CO2 produced from 45.0 L at STP (22.4 L/mol). From the chemical equation, it can be determined that 1 mole of CaCO3 produces 1 mole of CO2. Using the molar mass of CaCO3, we can then convert moles of CO2 to grams of CaCO3.
What is the grams of substance in 4.5 moles CaCO3?
Molar mass of CaCO3 = 66.1221g CaCO3/mole CaCO3. This means that 1 mole CaCO3 = 66.1221g CaCO3. To find the mass of 4.5 mole CaCO3, complete the following calculation: 4.5g CaCO3 X 1mol CaCO3/66.1221g CaCO3 = 0.068 mole CaCO3.
Calculate the number of grams in these quantities 125kg CaCO3?
Just 125 kg of CaCO3 contains 125000 grams of the substance. I kilogram of any substance is equivalent to 1000 grams of it.
what would be the volume of CO2 (at STP) produced from the complete reaction of 10 grams of CaCO2?
Assuming complete reaction, the molar mass of CaCO3 is approximately 100.09 g/mol. One mole of CaCO3 produces one mole of CO2. Therefore, 10 grams of CaCO3 will produce approximately 2.24 liters of CO2 at STP (22.4 L/mol).
How many grams of CaCO3 form if 155g of water reacts?
To determine the amount of CaCO3 formed, we need to know the molar ratio of water to CaCO3 in the chemical reaction. Without this information, it is not possible to calculate the amount of CaCO3 formed when 155g of water reacts.
What mass of water would a ton of calcium carbonate produce if it was fully reacted with hydrochloric acid?
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O 1 tonne = 2000 lbs ( I guess this is what you mean ) 2000 lbs CaCO3 (454 grams/1 lb)(1 mole CaCO3/100.09 grams)(1 mole H2O/1 mole CaCO3)(18.016 grams/1 mole H2O) = 163438.1856 grams of water ---------------------------- 163438.1856 grams (1 lb/454 grams) = 360 pounds of water -------------------------------
How many grams of calcium carbonate are needed to produce 45.0L of carbon dioxide at STP?
The balanced chemical equation for the reaction of calcium carbonate (CaCO3) to produce carbon dioxide (CO2) is: CaCO3 -> CaO + CO2. Using the ideal gas law, we can calculate the moles of CO2 produced from 45.0 L at STP (22.4 L/mol). From the chemical equation, it can be determined that 1 mole of CaCO3 produces 1 mole of CO2. Using the molar mass of CaCO3, we can then convert moles of CO2 to grams of CaCO3.
What is the grams of substance in 4.5 moles CaCO3?
Molar mass of CaCO3 = 66.1221g CaCO3/mole CaCO3. This means that 1 mole CaCO3 = 66.1221g CaCO3. To find the mass of 4.5 mole CaCO3, complete the following calculation: 4.5g CaCO3 X 1mol CaCO3/66.1221g CaCO3 = 0.068 mole CaCO3.
How many grams of oxygen will be required to reacts completely with 0.3 moles of aluminium?
999 g
Calculate the number of grams in these quantities 125kg CaCO3?
Just 125 kg of CaCO3 contains 125000 grams of the substance. I kilogram of any substance is equivalent to 1000 grams of it.
what would be the volume of CO2 (at STP) produced from the complete reaction of 10 grams of CaCO2?
Assuming complete reaction, the molar mass of CaCO3 is approximately 100.09 g/mol. One mole of CaCO3 produces one mole of CO2. Therefore, 10 grams of CaCO3 will produce approximately 2.24 liters of CO2 at STP (22.4 L/mol).
How many grams of H2O are made when 0.240 mol of methane reacts completely with oxygen?
The answer is 8,64 g.
If 1.25 grams of pure CaCO3 required 25.5 mL of a HCl solution for complete reation what is the molarity for the HCl solution?
When calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl) the reaction is:- CaCO3 + 2 HCl -> CaCl2 + H2O + CO2 The weight for 1 mole of calcium carbonate is the atomic weight in grams or 40.8 + 12.011 + 3 x 15.9994 = 100.8092 grams 1.25 grams of CaCO3 is therefore 1.25/100.8092 or .0124 mole There must have been twice this many moles of HCl in the solution or 0.0248 mole. The molarity of the solution is the number of moles per litre so with 0.0248 in 25.5 ml there must be .97 mole in 1 litre. The HCl solution must therefore be .97 molar
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
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)
When calcium carbonate decomposes according to the equation caco3-cao CO2 how many grams of CO2 are produced from the decomposition of 520 g of caco3?
To find the grams of CO2 produced from the decomposition of 520 g of CaCO3, we first need to calculate the molar mass of CaCO3, which is 100.09 g/mol. This means 520 g of CaCO3 is equal to 5.19 moles. From the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 5.19 moles of CaCO3 will produce 5.19 moles of CO2 which is equal to 235.10 g of CO2.
How many moles are there in 250 g of CaCO3?
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate
