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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.
What mass of caco3 is required to react with 100 ml of 2 m hcl solution?
To find the mass of CaCO3 required to react with 100 mL of 2 M HCl, you need to first calculate the number of moles of HCl using its molarity and volume. Then, use the balanced chemical equation to determine the mole ratio between HCl and CaCO3, allowing you to calculate the mass of CaCO3 needed.
How do you determine the number of kilograms of CaCO3 needed to neutralize 500 L of 1.40 M HCl?
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l) 500 L * (1.40 mol HCl/1 L HCl) * (1 mol CaCO3/2 mol HCl) * (100.09 g CaCO3/1 mol CaCO3) = 35031.5 g CaCO3 35031.5 g * (1 kg/1000 g) = 35.0315 kg Therefore, about 35.03 kilograms of calcium carbonate is needed to neutralize 500 liters of 1.40 M hydrochloric acid.
Calculate the number of moles in 27.50 grams CaCO3?
To calculate the number of moles in 27.50 grams of CaCO3, you need to divide the given mass by the molar mass of CaCO3. The molar mass of CaCO3 is approximately 100.09 g/mol. So, 27.50 grams divided by 100.09 g/mol gives you approximately 0.275 moles of CaCO3.
How many moles of CaCO3 is 73.4kg?
To find the number of moles in 73.4 kg of CaCO3, we first need to calculate the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol. Converting 73.4 kg to grams gives 73,400 g. Dividing 73,400 g by the molar mass of CaCO3 gives approximately 733 moles.
How many o atoms are present in 50gram of caco3?
There are 2 oxygen atoms in one molecule of CaCO3. To calculate the number of oxygen atoms in 50 grams of CaCO3, you first need to find the number of moles of CaCO3 using its molar mass. Then, multiply the number of moles by the number of atoms of oxygen per molecule of CaCO3 (2) to find the total number of oxygen atoms.
How do you determine the number of atoms in CaCO3?
To determine the number of atoms in CaCO3, you would first calculate the molar mass of CaCO3 (40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol)). Then, you would divide the given mass of CaCO3 by the molar mass to find the number of moles. Finally, you would use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to the number of atoms in CaCO3.
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.
What mass of caco3 is required to react with 100 ml of 2 m hcl solution?
To find the mass of CaCO3 required to react with 100 mL of 2 M HCl, you need to first calculate the number of moles of HCl using its molarity and volume. Then, use the balanced chemical equation to determine the mole ratio between HCl and CaCO3, allowing you to calculate the mass of CaCO3 needed.
How do you determine the number of kilograms of CaCO3 needed to neutralize 500 L of 1.40 M HCl?
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l) 500 L * (1.40 mol HCl/1 L HCl) * (1 mol CaCO3/2 mol HCl) * (100.09 g CaCO3/1 mol CaCO3) = 35031.5 g CaCO3 35031.5 g * (1 kg/1000 g) = 35.0315 kg Therefore, about 35.03 kilograms of calcium carbonate is needed to neutralize 500 liters of 1.40 M hydrochloric acid.
Calculate the number of moles in 27.50 grams CaCO3?
To calculate the number of moles in 27.50 grams of CaCO3, you need to divide the given mass by the molar mass of CaCO3. The molar mass of CaCO3 is approximately 100.09 g/mol. So, 27.50 grams divided by 100.09 g/mol gives you approximately 0.275 moles of CaCO3.
How many moles of CaCO3 is 73.4kg?
To find the number of moles in 73.4 kg of CaCO3, we first need to calculate the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol. Converting 73.4 kg to grams gives 73,400 g. Dividing 73,400 g by the molar mass of CaCO3 gives approximately 733 moles.
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 many grams of CaCO3 form if 155g of water reacts?
To determine the amount of CaCO3 formed, we need to know the molar ratio of water to CaCO3 in the chemical reaction. Without this information, it is not possible to calculate the amount of CaCO3 formed when 155g of water reacts.
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 formula units does 200 grams of Calcium Carbonate represent?
Well, darling, to find the number of formula units in 200 grams of Calcium Carbonate, you first need to calculate the molar mass of CaCO3. Once you do the math, you'll find that one mole of CaCO3 weighs 100.09 grams. So, 200 grams of CaCO3 represents 2 moles of the compound, which is equivalent to 2 times Avogadro's number (6.022 x 10^23) formula units. Voilà!
How many grams of calcium in 34.5 of Ca CO 3?
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.
