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250 grams CaCO3 (1 mole CaCO3/100.09 grams)
= 2.50 moles of calcium carbonate
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∙ 13y ago
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How much is 3 moles of marbles?
Marble is mainly CaCO3 so its one mole is equal to 100 g.
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
How many moles are in 10.10 g of calcium carbonate?
Calcium carbonate has the formula CaCO3 . Refer to the Periodic Table for atomic masses. 1 x Ca = 1 x 40 = 40 1 X C = 1 x 12 = 12 3 x O = 3 x 16 = 48 40 +12 +48 = 100 (The Mr : Reltive Molecular mass of CaCO3) Next use the wquation Moles = mass(g) / Mr Substituting Moles = 10.10 g / 100 moles = 0.101 moles.
Calculate the number of moles in 27.50 grams CaCO3?
Approximally 0.275 moles. The molar mass of Ca C O3 is ~ 40+12+3*16 = 100g/mol 27.50g = x mol *100g/mol 27.50g/(100g/mol) = x mol 0.275 g/(g/mol) = x mol 0.275 mol = x mol
How many liters of carbon dioxide can be obtained from the reaction of 906 grams of calcium carbonate with hydrochloric acid?
To determine the number of liters of carbon dioxide produced in this reaction, we need the balanced equation and the molar mass of carbon dioxide. The balanced equation is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O The molar mass of CO2 is 44.01 g/mol. First, we calculate the number of moles of CaCO3: 906 g / molar mass of CaCO3 = moles of CaCO3 Using the balanced equation, we see that the stoichiometric coefficient of CO2 is 1. This means that the number of moles of CO2 produced is equal to the number of moles of CaCO3. Finally, we convert moles of CO2 to liters using the ideal gas law: moles of CO2 x 22.4 L/mol = liters of CO2. Therefore, the number of liters of CO2 produced from 906 grams of CaCO3 can be calculated as follows: liters of CO2 = (906 g / molar mass of CaCO3) x 22.4 L/mol
How many moles are in 250 g iron (III) oxide?
250 g iron (III) oxide is equal to 1,565 moles.
How much is 3 moles of marbles?
Marble is mainly CaCO3 so its one mole is equal to 100 g.
How many grams of calcium carbonate are needed to produce 93.0 L of carbon dioxide at STP?
At STP one mole of any gas occupies 22.4 L. The equation for the reaction is CaCO3(S) - -> CaO (s) + CO2 (g). Moles of CO2 in 93.0 L is 93/22.4 = 4.15 moles, the ratio of moles of CaCO3 and CO2 is 1:1, therefore 4.15 moles of CaCO3 will be required. 1 mole of CaCO3 is equal to 100.087 g/mol. Therefore, 4 .15 moles will have 415.2 grams.
How many grams of calcium carbonate are needed to produce 95.0 L of carbon dioxide?
In 95.0 liters of CO2, there are 3.98 moles of CO2. This is determined from the molar mass of CO2 (44.010 g/mol) and that in 95.0 liters, there is .175 kg. (Density of CO2 is .001842 g/cm3) The stoichiometric equation to produce CO2 from CaCO3 is CaCO3 -> CaO + CO2. Because the equation is balanced with all coefficients being 1, it takes an equivalent number of moles of CaCO3 to produce the CO2. 3.98 moles of CaCO3 = 398 grams. (Molar weight of calcium carbonate is 100.1 g/mol)
How many moles are in 10.10 g of calcium carbonate?
Calcium carbonate has the formula CaCO3 . Refer to the Periodic Table for atomic masses. 1 x Ca = 1 x 40 = 40 1 X C = 1 x 12 = 12 3 x O = 3 x 16 = 48 40 +12 +48 = 100 (The Mr : Reltive Molecular mass of CaCO3) Next use the wquation Moles = mass(g) / Mr Substituting Moles = 10.10 g / 100 moles = 0.101 moles.
Calculate the volume of CO2 produced at STP from the decomposition of 152 g calcium carbonate caco3?
Molar mass CaCO3 = 100.087 g/mol Moles CaCO3 = 152 g / 100.087 = 1.52 the ratio between CaCO3 and CO2 is 1 : 1 so we get 1.52 moles of CO2 At STP p=1 ATM and T = 273 K V = nRT / p = 1.52 x 0.0821 x 273 /1 = 34.1L
How many grams are in 1.25 moles of calcium carbonate?
To answer this we must first find the molar mass of calcium carbonate. CaCO3Ca= 40.08gC=12.01gO= 16.00g (we have three oxygens so 16.00x3 is 48.00g)40.08+12.01+48.00= 100.09 gNow that we have the molar mass we can find how many grams there are:1.25 moles CaCo3 x (100.09 g CaCO3/ 1 mole CaCO3)= 125.11 grams CaCO3Therefore we'd have about 125 grams of CaCO3
How many atoms are present in 254g of platinum?
The standard atomic weight of platinum is 195,084 g.195,084 g-------------------------------1 mol250 g-------------------------------------x molx = 250/195,084 = 1,28 mol platinum
Calculate the number of moles in 27.50 grams CaCO3?
Approximally 0.275 moles. The molar mass of Ca C O3 is ~ 40+12+3*16 = 100g/mol 27.50g = x mol *100g/mol 27.50g/(100g/mol) = x mol 0.275 g/(g/mol) = x mol 0.275 mol = x mol
How many liters of carbon dioxide can be obtained from the reaction of 906 grams of calcium carbonate with hydrochloric acid?
To determine the number of liters of carbon dioxide produced in this reaction, we need the balanced equation and the molar mass of carbon dioxide. The balanced equation is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O The molar mass of CO2 is 44.01 g/mol. First, we calculate the number of moles of CaCO3: 906 g / molar mass of CaCO3 = moles of CaCO3 Using the balanced equation, we see that the stoichiometric coefficient of CO2 is 1. This means that the number of moles of CO2 produced is equal to the number of moles of CaCO3. Finally, we convert moles of CO2 to liters using the ideal gas law: moles of CO2 x 22.4 L/mol = liters of CO2. Therefore, the number of liters of CO2 produced from 906 grams of CaCO3 can be calculated as follows: liters of CO2 = (906 g / molar mass of CaCO3) x 22.4 L/mol
How many grams of KCl would you need to make 250 ml of a 0.5-M Solution w work?
Molar mass of KCl = 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol. A 0.5 M solution is required (0.5 mol/L or 0.5 moles per litre). 0.5 moles of KCl is 0.5 mol x 74.5 g/mol = 37.25 g. Dissolving this 37.25 g of KCl in a litre of water would give a 0.5 M solution. If 1 L or 1000 mL of 0.5 M solution contains 0.5 moles then 1 mL of the same concentration solution would contain 0.5/1000 moles and 250 mL would contain 250 x 0.5/1000 moles = 0.125 moles. 0.125 moles of KCl is 0.125 mol x 74.5 g/mol = 9.31 g.
How many moles does 3.80 g Zn have?
3,80 g Zn have 0,058 moles.
