This question is solved with the help of mole concept . 1 mole of anhydrous calcium carbonate weighs 40+12+48=100 gm . 1.25 mole of similar anhydrous calcium carbonate will be 100* 1.25 = 125 gm
1400 grams
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
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
Molar mass of calcium carbonate/calcium valence = 50,04345
Probably about 80%. This is what I have heard and learnt.
1400 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.
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.
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.)
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
For a partly ionically bonded compound such as calcium carbonate, the gram formula mass is substituted for a mole, which technically exists only for purely covalently bonded compounds. The gram formula mass for calcium carbonate is 100.09. Therefore, 200 grams constitutes 200/100.09 or 2.00 gram formula masses of calcium carbonate, to the justified number of significant digits.
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
Molar mass of calcium carbonate/calcium valence = 50,04345
Many antacids and calcium supplements contain 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
The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol. To calculate the percent mass of calcium, you need to divide the molar mass of calcium (40.08 g/mol) by the molar mass of calcium carbonate. This gives you a result of 0.4006, meaning that calcium constitutes approximately 40.06% of the mass of calcium carbonate.
Probably about 80%. This is what I have heard and learnt.