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.)
10
2,8 moles of calcium carbonate have 240,208 g.
5
9.03x10 23 ions
The answer is 19,288 g cacium carbonate.
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.
2,8 moles of calcium carbonate have 240,208 g.
6,863 grams of CaO
5
9.03x10 23 ions
1400 grams
Calcium has Wt 40, carbon is 12 and oxygen 16, so the MWt of calcium carbonate is 40+12+3x16=100. As this contains 40 calcium, calcium carbonate is 40% calcium. 40% of 418 is 167.2grams
0.720940834 grams
The answer is 19,288 g cacium carbonate.
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.
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.
caco3 (40)+(12)+3(16) =100/40 =2.5
201