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How many moles of CaCO3 is 73.4kg?
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 = 733molThere are 733 moles of calcium carbonate in a 73.4kg pure sample.
What is 1 Mole of CaCO3?
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".
How many grams of calcium carbonate are required to prepare 50.0 g of calcium oxide?
This is a mass stoichiometry problem. Start with the balanced equation: CaCO3 --> CaO + CO2. Do a conversion from 50g CaO to moles: 56g/1mol=50g/x, x=.9 moles. The equation is balanced as written, with all coefficients understood to be 1. So: .9 moles CaO means .9 moles CaCO3. Do another conversion from moles to grams: 100g/1mol=x/.9 moles. Solve for x to get 90 grams. (56g=molar mass of calcium oxide; 100g=molar mass of calcium carbonate.)
How many moles are in 10.10 g of calcium carbonate?
Calcium carbonate has the formula CaCO3 . Refer to the Periodic Table for atomic masses. 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 (The Mr : Reltive Molecular mass of CaCO3) Next use the wquation Moles = mass(g) / Mr Substituting Moles = 10.10 g / 100 moles = 0.101 moles.
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)
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 moles of CaCO3 is 73.4kg?
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 = 733molThere are 733 moles of calcium carbonate in a 73.4kg pure sample.
How do you convert 50 g of calcium carbonate into moles?
1. Calculate formula massCalcium carbonate has chemical formula CaCO3.Its formula mass is 40.1 + 12.0 + 3(16.0) = 100.12. Apply formula to calculate number of moles of CaCO3Amount of CaCO3= mass/formula mass= 50/100.1= 0.50mol
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
What is the percentage of limestone on earth?
Probably about 80%. This is what I have heard and learnt.
What is 1 Mole of CaCO3?
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".
How many grams of calcium carbonate are required to prepare 50.0 g of calcium oxide?
This is a mass stoichiometry problem. Start with the balanced equation: CaCO3 --> CaO + CO2. Do a conversion from 50g CaO to moles: 56g/1mol=50g/x, x=.9 moles. The equation is balanced as written, with all coefficients understood to be 1. So: .9 moles CaO means .9 moles CaCO3. Do another conversion from moles to grams: 100g/1mol=x/.9 moles. Solve for x to get 90 grams. (56g=molar mass of calcium oxide; 100g=molar mass of calcium carbonate.)
How many milligrams of calcium are present in 500 mg tablet of calcium carbonate?
The formula for calcium carbonate is CaCO3, and it has a gram formula mass of 100.09. The formula shows that each formula mass contains one calcium atom, which has a gram atomic mass of 40.08. Therefore, the mass fraction of calcium in calcium carbonate is 40.08/100.09, so that a 500 mg tablet of calcium carbonate contains 200 mg of calcium, to the justified number of significant digits.
Calculate the mass in grams of 2.5 moles in calcium carbonate?
For this you need the atomic (molecular) mass of CaCO3. Take the number of moles and multiply it by the atomic mass. Divide by one mole for units to cancel. CaCO3= 100.1 grams2.50 moles CaCO3 × (100.1 grams) = 250.25 grams CaCO3
How many moles are in 10.10 g of calcium carbonate?
Calcium carbonate has the formula CaCO3 . Refer to the Periodic Table for atomic masses. 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 (The Mr : Reltive Molecular mass of CaCO3) Next use the wquation Moles = mass(g) / Mr Substituting Moles = 10.10 g / 100 moles = 0.101 moles.
What mass of calcium oxide is produced by heating 25 moles of calcium carbonate?
1400 grams
What percentage of an egg's shell is made of calcium carbonate?
you need to do what is called a back-titration, by reacting the seashell (ground up very fine) with an acid, then titrating the solution with a base. the data can then be used to find the moles of CaCo3, and from there you can go to grams. Divide grams CaCo3 by the mass of seashell used, then multiply by 100 to find the percentage
