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The formula mass of calcium carbonate, CaCO3 is 40.1 + 12.0 + 3(16.0) = 100.1Note that 73.4kg is equivalent to 73400g.
Amount of CaCO3 = mass of pure sample/molar mass = 73400/100.1 = 733mol
There are 733 moles of calcium carbonate in a 73.4kg pure sample.
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∙ 9y ago
Wiki User
∙ 10y ago2.43 mol CaCO3
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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
How many moles of oxygen atoms are in 4.25 moles of calcium carbonate the chief constituent of seashells?
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
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 much is 3 moles of marbles?
Marble is mainly CaCO3 so its one mole is equal to 100 g.
How many liters of carbon dioxide can be obtained from the reaction of 906 grams of calcium carbonate with hydrochloric acid?
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
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
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
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 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 atoms are in 4.25 moles of calcium carbonate the chief constituent of seashells?
Calcium Carbonate (CaCO3)I mol CaCO3 contains 3 mol Oxygen atomsso 4.25 mol CaCO3 will have 12.75 mol Oxygen Atoms.
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.
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 calcium carbonate are in 1.719 grams?
Calcium carbonate, CaCO3 has formula mass of 40.1+12.0+3(16.0) = 100.1Amount of CaCO3 = 1.719/100.1 = 0.0172molThere are 0.0172 moles of calcium carbonate in a 1.719 gram pure sample.
How many grams are in 1.25 moles of calcium carbonate?
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3
How much is 3 moles of marbles?
Marble is mainly CaCO3 so its one mole is equal to 100 g.
How many grams of calcium carbonate are needed to produce 49.0 L of carbon dioxide at STP?
n=(1atm)(49L)/(0.0821)(273.15)=2.19 moles2.19 moles x 44g of CO2 =96.36g of CO22.19 moles x 100g of CaCO3= 219g of CaCO3
How many moles of calcium atoms are 2.5 moles of calcium carbonate CaCO3?
The formula given shows that each formula unit or mole contains one calcium atom; therefore, 2.5 moles of calcium chloride contains 2.5 moles of calcium atoms.
