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50.05
The molar mass of CaCO3 is approximately 100.09 g/mol. Therefore, the mass of 0.5 moles of CaCO3 would be 50.045 g.
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What is the grams of substance in 4.5 moles CaCO3?
To find the grams of CaCO3 in 4.5 moles, you would first calculate the molar mass of CaCO3 (40.08 g/mol for Ca, 12.01 g/mol for C, and 16.00 g/mol for O), which totals 100.09 g/mol. Then, multiply this molar mass by the number of moles (4.5) to get the grams of CaCO3. So, 4.5 moles of CaCO3 is equivalent to 450.405 grams.
What does the molar mass of C7H16 and the molar mass of CaCO3 have the same number of?
Both C7H16 and CaCO3 have a molar mass of approximately 116 grams per mole.
How many moles are there in 250 g of CaCO3?
To find the number of moles in 250 g of CaCO3, divide the given mass by the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol. So, 250 g ÷ 100.09 g/mol ≈ 2.50 moles of CaCO3.
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.
What mass of lime can be produced from 1.5 10 3 kg of limestone?
To calculate the mass of lime produced, we need to consider the chemical equation for the reaction: CaCO3 (limestone) -> CaO (lime) + CO2 The molar mass of CaCO3 is 100.09 g/mol and that of CaO is 56.08 g/mol. In this reaction, 1 mole of CaCO3 produces 1 mole of CaO. First, convert 1.5 x 10^3 kg of limestone to grams, then divide by the molar mass of CaCO3 to find the number of moles. Finally, multiply the moles of CaCO3 by the molar ratio of CaCO3 to CaO to find the mass of lime produced.
What is the grams of substance in 4.5 moles CaCO3?
To find the grams of CaCO3 in 4.5 moles, you would first calculate the molar mass of CaCO3 (40.08 g/mol for Ca, 12.01 g/mol for C, and 16.00 g/mol for O), which totals 100.09 g/mol. Then, multiply this molar mass by the number of moles (4.5) to get the grams of CaCO3. So, 4.5 moles of CaCO3 is equivalent to 450.405 grams.
What does the molar mass of C7H16 and the molar mass of CaCO3 have the same number of?
Both C7H16 and CaCO3 have a molar mass of approximately 116 grams per mole.
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?
To calculate the percent of CaCO3 in the limestone, first determine the molar mass of CaCO3 (100.09 g/mol). Next, calculate the moles of CO2 produced from the reaction (0.44 g CO2 / 44.01 g/mol). Since 1 mole of CaCO3 produces 1 mole of CO2, the moles of CaCO3 in the limestone can be calculated. Finally, determine the mass percent of CaCO3 in the limestone (mass of CaCO3 / total mass of limestone).
How do you convert alkalinity as HCO3 to CaCO3?
To convert alkalinity (HCO3) to CaCO3, you need to use the molar mass ratio. For every mole of bicarbonate (HCO3), you have one mole of carbonate (CO3) in CaCO3. So, to convert, you can multiply the HCO3 concentration by a factor of 50.04 (molar mass of CaCO3/molar mass of HCO3).
How many moles are there in 250 g of CaCO3?
To find the number of moles in 250 g of CaCO3, divide the given mass by the molar mass of CaCO3. The molar mass of CaCO3 is 100.09 g/mol. So, 250 g ÷ 100.09 g/mol ≈ 2.50 moles of CaCO3.
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.
How many grams of calcium carbonate are needed to produce 55.0 L of carbon dioxide at STP?
The balanced chemical equation for the production of carbon dioxide from calcium carbonate is CaCO3 → CaO + CO2. From this equation, it can be seen that one mole of CaCO3 produces one mole of CO2. At STP, one mole of any gas occupies 22.4 L. Therefore, 55.0 L of CO2 corresponds to 55.0/22.4 = 2.46 moles of CO2. Since one mole of CaCO3 produces one mole of CO2, 2.46 moles of CO2 would require 2.46 moles of CaCO3. Finally, the molar mass of CaCO3 is approximately 100 g/mol, so 2.46 moles of CaCO3 would be 2.46 * 100 = 246 grams of CaCO3 are needed.
What mass of lime can be produced from 1.5 10 3 kg of limestone?
To calculate the mass of lime produced, we need to consider the chemical equation for the reaction: CaCO3 (limestone) -> CaO (lime) + CO2 The molar mass of CaCO3 is 100.09 g/mol and that of CaO is 56.08 g/mol. In this reaction, 1 mole of CaCO3 produces 1 mole of CaO. First, convert 1.5 x 10^3 kg of limestone to grams, then divide by the molar mass of CaCO3 to find the number of moles. Finally, multiply the moles of CaCO3 by the molar ratio of CaCO3 to CaO to find the mass of lime produced.
What mass of water would a ton of calcium carbonate produce if it was fully reacted with hydrochloric acid?
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O 1 tonne = 2000 lbs ( I guess this is what you mean ) 2000 lbs CaCO3 (454 grams/1 lb)(1 mole CaCO3/100.09 grams)(1 mole H2O/1 mole CaCO3)(18.016 grams/1 mole H2O) = 163438.1856 grams of water ---------------------------- 163438.1856 grams (1 lb/454 grams) = 360 pounds of water -------------------------------
How many Ca atoms are found in 0.50 moles of CaCO3?
0.50 moles CaCO3 (1 mole Ca/1 mole CaCO3)(6.022 X 1023/1 mole Ca)= 3.0 X 1023 atoms of calcium===================
How many moles of oxygen are in one mole of calcium carbonate?
There are three moles of oxygen atoms in one mole of calcium carbonate (CaCO3). This is because there are three oxygen atoms in each molecule of calcium carbonate.
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.
