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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 calcium carbonate are needed to produce 71.0 L of carbon dioxide at STP?
Assuming complete reaction, the molar volume of gas at STP is 22.4 L. Therefore, 71.0 L of CO2 corresponds to 71.0/22.4 = 3.17 moles of CO2. From the balanced chemical equation for the reaction of calcium carbonate (CaCO3) -> CaO + CO2, 1 mole of CaCO3 produces 1 mole of CO2. So, 3.17 moles of CO2 requires 3.17 moles of CaCO3. The molar mass of CaCO3 is 100.1 g/mol, so 3.17 moles of CaCO3 would be 3.17 * 100.1 = 317.6 grams of CaCO3.
How many grams of calcium in 34.5 of Ca CO 3?
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g 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 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.
How many Ca atoms are found in 0.50 moles of CaCO3?
0.50 moles CaCO3 (1 mole Ca/1 mole CaCO3)(6.022 X 1023/1 mole Ca)= 3.0 X 1023 atoms of calcium===================
How many moles of oxygen are in one mole of calcium carbonate?
1 mole CaCO3 (3 mole O/1 mole CaCO3) = 3 moles oxygen
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 much CaCO3 could be decomposed by 90 000 kj of heat energy?
delta H = (delta Ho f products) - (delta Ho f reactants) = ((1 mole CaO)(-636.6 kJ/mole) + (1 mole CO2)(-393.5 kJ/mole)) - (1 mole CaCO3)(-1206.9 kJ/mole) = -1030.1 kJ - (-1206.9 kJ) = +176.8 kJ This reaction is highly endothermic (positive delta H) at 25 C.
How many grams of calcium carbonate are needed to produce 71.0 L of carbon dioxide at STP?
Assuming complete reaction, the molar volume of gas at STP is 22.4 L. Therefore, 71.0 L of CO2 corresponds to 71.0/22.4 = 3.17 moles of CO2. From the balanced chemical equation for the reaction of calcium carbonate (CaCO3) -> CaO + CO2, 1 mole of CaCO3 produces 1 mole of CO2. So, 3.17 moles of CO2 requires 3.17 moles of CaCO3. The molar mass of CaCO3 is 100.1 g/mol, so 3.17 moles of CaCO3 would be 3.17 * 100.1 = 317.6 grams of CaCO3.
How many grams of calcium in 34.5 of Ca CO 3?
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.
How many moles of carbon are in 3.5 moles of calcium carbonate?
3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles
When a 1.25-gram sample of limestone was dissolved in acid 0.44 gram of CO2 was generated if the rock contained no carbonate other than CaCO3 what was the percent of CaCO3 by mass in the limestone?
CaCO3 +2HCl ------------> CaCl2 + CO2 + H2O number of moles of CO2 in .44 grams = .44/ 44 = .01 From equation it is clear that 1 mole of CO2 is produced from CaCO3 = 1 mole .01 mole of CO2 is formed from CaCO3 = .01 mole Weight of .01 mole of CaCO3 is = .01mole *100 g/mole = 1 gram weight % of CaCO3 is = 1*100/ 1.25 = 80 % w/w I've post my answer, so why don't you show that answer here with the question. It's fare. I must be informed about my answer weather it is right or wrong. Please inform me at amitmahalwar@yahoo.com
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 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.
Write a Balanced equation between HCl and CaCO3?
2 HCl + CaCO3 -> CaCl2 + H2O + CO2 In words, two moles of hydrochloric acid reacts with one mole of calcium carbonate to yield one mole of calcium chloride, one mole of water, and one mole of carbon dioxide.
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
