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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".
Wiki User
∙ 14y ago
Wiki User
∙ 13y agoCalcium carbonate (CaCO3)
Using theAtomic Masses from the Periodic Table
1 x Ca = 1 x 40 = 40
1 x C = 1 x 12 = 12
3 x O = 3 x 16 = 48
40 + 12 + 48 = 100 (Mr of CaCO3)
moles = mass / Mr
mass = moles x Mr
mass = 1 mole x 100 = 100grams.
Wiki User
∙ 10y agoCalcium carbonate, CaCO3 has formula mass of 40.1+12.0+3(16.0) = 100.1
Amount of CaCO3 = 62.4/100.1 = 0.623mol
There are 0.623 moles of calcium carbonate in a 62.4 gram pure sample.
Wiki User
∙ 10y agoThe formula mass of calcium chloride, CaCl2 is 40.1+2(35.5) = 111.1
Amount of calcium chloride = 10/111.1 = 0.090mol
There are 0.090 moles of calcium chloride in a 10 gram pure sample.
Wiki User
∙ 14y agon:number of moles
m:mass
M:Molar mass
Molar Mass of Calcium = 40
n=m/M
n=10g/40
n=o.25mol
Wiki User
∙ 11y ago100.087 g/mol (from wikipedia)
Molecular formula =
CaCO3
Ca = 40
C = 12
O = 16
Molar mass = 40 + 12 + 3(16)
= 100
Wiki User
∙ 6y agoThe molar mass of calcium carbonate is 100,0869 g.
Wiki User
∙ 6y ago100,0869 g/mol
Wiki User
∙ 11y ago6.02*1023
Anonymous
mole ratio of CaCO3 TO Ca
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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 in 34.5 of Ca CO 3?
34.5 g CaCO3 (1 mole CaCO3/100.09 g)(1 mole Ca/1 mole CaCO3)(40.08 g/1 mole Ca) = 13.8 grams calcium ===============
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
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 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
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 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?
34.5 g CaCO3 (1 mole CaCO3/100.09 g)(1 mole Ca/1 mole CaCO3)(40.08 g/1 mole Ca) = 13.8 grams calcium ===============
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 o atoms are present in 50gram of caco3?
50 grams CaCO3 (1 mole CaCO3/100.09 grams)(6.022 X 1023/1 mole CaCO3) = 3.0 X 1023 atoms of calcium carbonate =============================
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 -------------------------------
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
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 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.
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 93.0 L of carbon dioxide at STP?
At STP one mole of any gas occupies 22.4 L. The equation for the reaction is CaCO3(S) - -> CaO (s) + CO2 (g). Moles of CO2 in 93.0 L is 93/22.4 = 4.15 moles, the ratio of moles of CaCO3 and CO2 is 1:1, therefore 4.15 moles of CaCO3 will be required. 1 mole of CaCO3 is equal to 100.087 g/mol. Therefore, 4 .15 moles will have 415.2 grams.
