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0.89 moles of CaCl2 is equal to 98.8g.
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∙ 8y ago
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What is the molar weight of CaCl2?
The molar mass of CaCN2 = 80.1021 g/mol
Determine the number of moles of Cl in 1.9 mol CaCl2?
Ca-1(40.08)g=40.8g/mol of Ca Cl-2(35.45)g=70.9g/mol of Cl2 40.08g+70.9g=110.98g/mol CaCl2 110.98g CaCl2/1mol * 1.9mol =210.86 g of CaCl2
Find the mass of 0.89 mol of CaCl2?
Ca + Cl + Cl = 40.078 +35.453 + 35.453 = 110.984 .89 / 110.984 = 0.008019
A solution containing 14 grams of AgNo3 is added to a solution containing 4.83 g of CaCl2. Find the mass of the precipitate produced.?
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
What is the molality of a solution that has 6 mol of CaCl2 in 3 kg of water?
2 m
What is the gram fromula mass of CaCl2?
The formula mass of (anhydrous) CaCl2 is: 110,99 g/mol;as dihydrate CaCl2.2H2O it is: 147,01 g/mol
What is the molar weight of CaCl2?
The molar mass of CaCN2 = 80.1021 g/mol
Determine the number of moles of Cl in 1.9 mol CaCl2?
Ca-1(40.08)g=40.8g/mol of Ca Cl-2(35.45)g=70.9g/mol of Cl2 40.08g+70.9g=110.98g/mol CaCl2 110.98g CaCl2/1mol * 1.9mol =210.86 g of CaCl2
How many moles are in 1 g of Cl2?
# of Moles = Mass in grams divided by Molar Mass =5o divided by (cl x 2) =50 divided by 71 =0.704 moles use: 1 mol = Mr in grams that is 35.5x2 g of Cl2 = 1 mol 71g of Cl2 = 1 mol therefore 50g of Cl2 = (1/71) x 50 =0.704 mol
How would you make 450 mL of a 25 M calcium chloride solution?
You'll need: 0.450 (L) * 25 (mol/L) = 11.25 mol CaCl2 and add water to it up to 450 mL.To be weighted:11.25 (mol CaCl2) * [40.08 + 2*35.45](g/mol CaCl2) = 1491.736 g = 1500 g =1.5 kg CaCl2 (CaCl2 as dry substance!, not hydrated)
What is the molality of a solution that has 6 mol of CaCl2 in 3km of water?
2 m is the molarity of a solution that has 6 mol of CaCl2 in 3 km of water.
What is the mass of 4.7910-2 mol cacl2ything?
What is cacl2ything? Molar mass of CaCl2 is 111 g/mol Best guess for an answer to your confusing question is 4.79 x 10-2 mole x 111 g/mole = 5.3169 grams
Find the mass of 0.89 mol of CaCl2?
Ca + Cl + Cl = 40.078 +35.453 + 35.453 = 110.984 .89 / 110.984 = 0.008019
A solution containing 14 grams of AgNo3 is added to a solution containing 4.83 g of CaCl2. Find the mass of the precipitate produced.?
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
How many grams of CaCl2 are needed to prepare 125 ml of a 1.50 M solution?
Molarity = moles of solute/volume of solution 1.50 M = X Moles/125ml = 187.5 millimoles, which is 0.1875 moles The molar mass of CaCl2 = 110.98 grams 0.1875 moles CaCl2 (110.98g/1mol) = 20.8 grams needed
What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
What is the molality of a solution that has 6 mol of cacl2 in 3kg of water?
2m
