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What volume of CO2 forms from 5.25 grams of CaCo3 at 103killipascals and 25 degrees Celsius?
Wiki User
∙ 10y ago
The volume of carbon dioxide is 8,4 L at oC.
Wiki User
∙ 6y ago
To calculate the volume of CO2 formed, first determine the moles of CaCO3 using its molar mass. Then, calculate the moles of CO2 produced assuming all the CaCO3 is converted to CO2. Finally, use the ideal gas law (PV = nRT) to find the volume of CO2 at 103 kPa and 25°C. Be sure to convert Celsius to Kelvin (273 + 25).
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∙ 10y ago1.26
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Formula of limestone?
The formula for limestone is CaCO3, which represents calcium carbonate. It is a sedimentary rock that is composed mainly of the mineral calcite.
Is caco3 insoluble when react with water?
caco3 is solid form. its density is high so caco3 is not soluble in water. but co2 mix in water befuor the mix in caco3 . so caco3 is soluble after mixing the co2 .
What is the balanced equation for caCO3 2HCI?
CaCO3 + 2HCl --> CaCl2 + H2O + CO2
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.
Whats the limestone heating reaction?
When limestone (calcium carbonate) is heated, it undergoes thermal decomposition to produce calcium oxide (quicklime) and carbon dioxide gas. The chemical equation for this reaction is: CaCO3 (s) → CaO (s) + CO2 (g).
What is the volume of CaCO3 if you have a molarity of 75.5 and 15 g CaCO3 What is the answer and how did you get it?
Calcium carbonate is not soluble in water.
What volume of a 0.10 mol dm-3 solution of HCl is required to react with 10.0 g of CaCO3?
First, calculate the number of moles of CaCO3 using its molar mass. Then, determine the ratio between HCl and CaCO3 according to the balanced chemical equation. Finally, use the concentration of the HCl solution to find the required volume using the formula: volume = moles/concentration.
CaCO3 -- CO2 CaO how many liters of CO2 at 1.03 ATM and a temperature of 950 degrees celsius will be produced if there is 1.00 kg of CaCO3 and what are the calculations to solve it or formulas?
To solve this problem, you first need to calculate the number of moles of CaCO3 in 1.00 kg. Then, use the stoichiometry of the balanced chemical equation to determine the number of moles of CO2 produced. Finally, apply the ideal gas law to convert moles of CO2 to liters using the given conditions (1.03 atm and 950 degrees Celsius). The ideal gas law equation is: PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature.
When calcium carbonate is heated strongly carbon dioxide gas is released CaCO3CaO CO2 what is the volume of CO2 measured stpis produced if 15.2 grams of caco3 is heated?
The molar mass of CaCO3 is 100.09 g/mol, and the molar mass of CO2 is 44.01 g/mol. By using stoichiometry, you can calculate that 15.2 grams of CaCO3 would produce 6.51 grams of CO2. Using the ideal gas law, you can then convert the mass of CO2 to volume using its molar volume at STP (22.4 L/mol). The volume of CO2 produced would be around 3.32 liters.
What conditions of temperature and pressure would get the most decompomposition of CaCo3?
Calcium carbonate (CaCO3) decomposes into calcium oxide (CaO) and carbon dioxide (CO2) at high temperatures. To maximize decomposition, high temperatures above 800 degrees Celsius and low pressures are ideal. These conditions will promote the breakdown of CaCO3 into its components.
what would be the volume of CO2 (at STP) produced from the complete reaction of 10 grams of CaCO2?
Assuming complete reaction, the molar mass of CaCO3 is approximately 100.09 g/mol. One mole of CaCO3 produces one mole of CO2. Therefore, 10 grams of CaCO3 will produce approximately 2.24 liters of CO2 at STP (22.4 L/mol).
Which of the following salts has the greatest solubility in water at 25 degrees Celsius?
The salt with the greatest solubility in water is sodium chloride (NaCl). Sodium chloride is highly soluble in water due to its strong ionic bonds that readily break apart in water molecules, allowing it to dissociate into Na+ and Cl- ions.
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 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.
Formula of limestone?
The formula for limestone is CaCO3, which represents calcium carbonate. It is a sedimentary rock that is composed mainly of the mineral calcite.
How is density a size-independent property?
Purely from a definitional sense, density is size independent because it depends on the substance in question, not how much of it you have. From a mathematical sense, density is mass/volume. If the mass is equally distributed throughout the volume, then half of the volume, for example would have half the mass, preserving the same value of density.
How many grams of calcium carbonate are needed to produce 71.0 L of carbon dioxide at STP?
Assuming complete reaction, the molar volume of gas at STP is 22.4 L. Therefore, 71.0 L of CO2 corresponds to 71.0/22.4 = 3.17 moles of CO2. From the balanced chemical equation for the reaction of calcium carbonate (CaCO3) -> CaO + CO2, 1 mole of CaCO3 produces 1 mole of CO2. So, 3.17 moles of CO2 requires 3.17 moles of CaCO3. The molar mass of CaCO3 is 100.1 g/mol, so 3.17 moles of CaCO3 would be 3.17 * 100.1 = 317.6 grams of CaCO3.
