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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.
0,2748 mol
equation used: n=m/M
n is amount of substance (in mole)
m is mass of CaCO3 (27.50 g)
M is molar mass (for CaCO3 100,086 g/mol)
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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.
How do you solve for how many grams are in 2.38 moles of calcium carbonate?
To convert moles to grams, you need to use the molar mass of calcium carbonate (CaCO3). The molar mass of CaCO3 is approximately 100.1 g/mol. To calculate the grams in 2.38 moles of CaCO3, you would multiply the number of moles (2.38) by the molar mass (100.1 g/mol), which gives you approximately 238 grams.
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 CaCl2 can be obtained if 14.6 g HCl is allowed to react with excess CaCO3?
The balanced chemical equation for the reaction between HCl and CaCO3 is: 2HCl + CaCO3 → CaCl2 + H2O + CO2. The molar ratio between HCl and CaCl2 is 2:1. Calculate the number of moles of HCl from 14.6 g, then use the mole ratio to find the moles of CaCl2. Finally, convert moles of CaCl2 to grams.
How may grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide?
To calculate the grams of calcium carbonate needed, you first need to determine the moles of carbon dioxide produced (using the ideal gas law). Then, since 1 mole of CO2 is produced for every mole of CaCO3 consumed, you can convert moles of CO2 to moles of CaCO3. Finally, use the molar mass of CaCO3 to convert moles to grams.
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.
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.
How do you solve for how many grams are in 2.38 moles of calcium carbonate?
To convert moles to grams, you need to use the molar mass of calcium carbonate (CaCO3). The molar mass of CaCO3 is approximately 100.1 g/mol. To calculate the grams in 2.38 moles of CaCO3, you would multiply the number of moles (2.38) by the molar mass (100.1 g/mol), which gives you approximately 238 grams.
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.
Calculate the mass in grams of 2.5 moles in calcium carbonate?
The molar mass of CaCO3 (calcium carbonate) is 100.09 g/mol. To convert moles to grams, multiply the number of moles by the molar mass. Therefore, the mass of 2.5 moles of calcium carbonate is 250.225 grams.
How many grams of CaCl2 can be obtained if 14.6 g HCl is allowed to react with excess CaCO3?
The balanced chemical equation for the reaction between HCl and CaCO3 is: 2HCl + CaCO3 → CaCl2 + H2O + CO2. The molar ratio between HCl and CaCl2 is 2:1. Calculate the number of moles of HCl from 14.6 g, then use the mole ratio to find the moles of CaCl2. Finally, convert moles of CaCl2 to grams.
How may grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide?
To calculate the grams of calcium carbonate needed, you first need to determine the moles of carbon dioxide produced (using the ideal gas law). Then, since 1 mole of CO2 is produced for every mole of CaCO3 consumed, you can convert moles of CO2 to moles of CaCO3. Finally, use the molar mass of CaCO3 to convert moles to grams.
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 do you convert 50 g of calcium carbonate into moles?
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
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
To determine the number of grams of calcium carbonate needed to produce 95.0 L of carbon dioxide, you would first need to calculate the moles of carbon dioxide produced using the ideal gas law. Then, use the stoichiometry of the balanced chemical equation between calcium carbonate and carbon dioxide to convert moles of CO2 to moles of CaCO3, and finally to grams of CaCO3.
When calcium carbonate decomposes according to the equation caco3-cao CO2 how many grams of CO2 are produced from the decomposition of 520 g of caco3?
To find the grams of CO2 produced from the decomposition of 520 g of CaCO3, we first need to calculate the molar mass of CaCO3, which is 100.09 g/mol. This means 520 g of CaCO3 is equal to 5.19 moles. From the balanced chemical equation, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 5.19 moles of CaCO3 will produce 5.19 moles of CO2 which is equal to 235.10 g of CO2.
How many grams of calcium carbonate are needed to produce 93.0 L of carbon dioxide at STP?
Using the molar volume of a gas at STP (22.4 L/mol), we can calculate that 93.0 L of CO2 is equivalent to 93.0/22.4 = 4.15 moles of CO2. Since the balanced chemical equation for the reaction where calcium carbonate produces CO2 is CaCO3 => CaO + CO2, it can be seen that 1 mole of CaCO3 produces 1 mole of CO2. Therefore, 4.15 moles of CO2 would require 4.15 moles of CaCO3. Finally, using the molar mass of CaCO3 (100.09 g/mol), we can calculate that 4.15 moles of CaCO3 is equivalent to 4.15 * 100.09 = 415.84 grams of CaCO3. So, 415.84 grams of calcium carbonate are needed to produce 93.0 L of carbon dioxide at STP.
