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What mass of caco3 is required to react with 100 ml of 2 m hcl solution?
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∙ 12y ago
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∙ 12y ago
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What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
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 is the mass of HCl that would be required to completely react with 5.2 grams of Mg?
69
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 does the molar mass of C7H16 and the molar mass of CaCO3 have the same number of?
342.3
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
2 HCl + CaCO3 = CaCl2 + CO2 + H2O Therefore it takes twice as many moles of HCl to react with CaCO3 Mol CaCO3 = mass / formulae weight = 10 / 40.08 + 14.01 + 3(16.00) = 10 / 102.09 = 0.0980 mol Mol HCl = 2 x 0.0980 = 0.1959 mol Volume HCl = mol / c = 0.1959 / 0.10 = 1.9591 L Volume required to react of HCl = 1.9591 L
What is the mass of CaCO3?
Molar mass of CaCO3 = 100.0869 g/mol
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 lead(II) nitrate is required to react with 370-g NaOH?
The mass of lead(II) nitrate required to react with 370 g NaOH is 1 531,9 g.
How much 2.5 molarity of HNO3 is required to react the CaCO3?
This depends on the mass of calcium carbonate.
The molar mass of CaCO3 is greater than the molar mass of CaNO32?
Molar mass of CaCO3 = 100.0869 g/molMolar mass of CaNO32 = 566.0655 g/mol
What are the signs of chemical change if you combine a steel to the solution?
This depends on the nature of solution; when the sample react with an acid the mass decrease and the solution become colored.
What is the mass of HCl that would be required to completely react with 5.2 grams of Mg?
69
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 many grams of magnesium hydroxide required to make 1M of the solution?
the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g
What does the molar mass of C7H16 and the molar mass of CaCO3 have the same number of?
342.3
What mass of oxygen (O2) is required to react completely with 86.17 grams of C6H14?
The answer is 152 g oxygen.
