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The formula of calcium carbonate is CaCO3. Therefore, its molar mass is the sum of the atomic masses of calcium and carbon and three times the Atomic Mass of oxygen:
40.078 + 12.011 + 3(15.999) = 100.086. The number of moles in 10 g is 10 divided by the molar mass = 0.10 moles, to the number of significant digits justified by "10 g".
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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 mole h2o have in this reaction caco3 plus 2hcl---cacl2 co2 h2o?
Molar ratio's in this balanced equation: 1 + 2 --> 1 + 1 + 1 (H2O) respectively
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.
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
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.
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
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
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 -------------------------------
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 mole h2o have in this reaction caco3 plus 2hcl---cacl2 co2 h2o?
Molar ratio's in this balanced equation: 1 + 2 --> 1 + 1 + 1 (H2O) respectively
How many o atoms are present in 50gram of caco3?
There are 2 oxygen atoms in one molecule of CaCO3. To calculate the number of oxygen atoms in 50 grams of CaCO3, you first need to find the number of moles of CaCO3 using its molar mass. Then, multiply the number of moles by the number of atoms of oxygen per molecule of CaCO3 (2) to find the total number of oxygen atoms.
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.
Calculate the volume of CO2 produced at STP from the decomposition of 152 g calcium carbonate caco3?
Molar mass CaCO3 = 100.087 g/mol Moles CaCO3 = 152 g / 100.087 = 1.52 the ratio between CaCO3 and CO2 is 1 : 1 so we get 1.52 moles of CO2 At STP p=1 ATM and T = 273 K V = nRT / p = 1.52 x 0.0821 x 273 /1 = 34.1L
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.
