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KCl
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Help with stoichiometry show steps How many grams of KClO3 must be decomposed to yield 30 grams of oxygen?
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
What is the molarity of the following solution 1.0 moles of KCl in 500mL of solution?
You need to bring this to a litre. If there is 1 mole in 500ml (using ratio), there will be 2 moles in a litre. So your molarity is 2.
How many moles of potassium hydroxide would be required to completely neutralize 0.100 mole of hydrochloric acid?
The same molar amount, 0.100mol KOH. The reaction is in a 1:1 stoichiometric ratio: HCl + KOH --> KCl + H2O.
What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas?
Balanced equation and potassium limits and drives the reaction.2K + Cl2 -> 2KCl6.75 grams K (1 mole K/39.10 grams)(2 mole KCl/2 mole K)(74.55 grams /1 mole KCl)= 12.9 grams potassium chloride produced==============================
What is the mole ratio of salt to water in NA2SO4.10H2O?
An anhydrous sals hasn't water.
How can you prepare 0.01M KCL solution in 1liter?
Dissolve 0.01 mole of KCl in upto 1 Liter. 0.01 mole = 0.01 (mol) * MKCl (g/mol KCl) = (0.01*MKCl) gram KCl per Liter
How many mols in 74g of KCl?
1 mole.
How do you prepare 3.5M KCl solution?
3.5M means 3.5 moles of KCl. 1 mole is the combined molecular weight of the compound per litre. Molecular weight of K (potassium) = 39.10g Molecular weight of Cl (chlorine) = 35.45g So molecular weight of KCl = (39.10 + 35.45) = 74.55g That means that 1 mole of KCL = 74.55 grams per litre If 1 mole of KCL contains 74.55g then 3.5M of KCL will contain 74.55g x 3.5 and so 3.5M of KCL = 260.925g/L
How does the experimental mole ratio compare with the theoretical mole ratio?
the experimental mole ratio has a bigger penis
What is the molarity of a solution made by dissolving 4.88 g of KCl in 423 mL of solution?
Need mole KCl first. 4.88 grams KCl (1 mole KCl/74.55 grams) = 0.06546 moles KCl =======================now, Molarity = moles of solute/Liters of solution ( 423 ml = 0.423 Liters ) Molarity = 0.06546 moles KCl/0.423 Liters = 0.155 M KCl ------------------
How many moles are in 74 g of KCl?
1 mole
Help with stoichiometry show steps How many grams of KClO3 must be decomposed to yield 30 grams of oxygen?
2 KClO3 -> KCL + 3O2 Molar weight of O2 = 32 grams/mole (so close it doesn't matter) 30 grams/32grams/mole = 0.9375 moles Molar weight of KCL = 39+35.5 = 74.5 grams/mole (Want more accuracy? Do it yourself?) now if we have 3 moles of O2 then we have 2 moles of KCl. If we have one mole of O2 then we have 2/3 moles of KCL What ever moles we have of O2 we must multiply it by 2/3 to get the moles of KCl So we have 0.9375moles of O2 x 2/3 = 0.625 moles of KCl So 0.625 moles of KCl x 74.5 grams/mole KCl = 46.5625 grams KCl
A solution was prepared by dissolving 40.0 g of KCl in 225 g of water. Calculate the mole fraction of KCl in the solution?
18%
What is the molarity of the following solution 1.0 moles of KCl in 500mL of solution?
You need to bring this to a litre. If there is 1 mole in 500ml (using ratio), there will be 2 moles in a litre. So your molarity is 2.
How many grams of KCl are needed to make 100 ml of a 3.5M solution?
In a 3.4 M solution, there are 3.4 moles per liter. If you want to make 3 liters of solution, you'll need 3 liter * 3.4 moles/liter = 10.2 moles The molar mass of KCl is 39.098 g/mole K + 35.453 g/mole Cl = 74.551 g/mole KCl To get the number of grams, multiply the number of moles by the molar mass: 10.2 moles * 74.551 g/mole KCl = 760.4202 g = 0.760 kg
How many grams of KCl are required to prepare 2.25L of a 0.75m KCl solution?
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
How many moles of potassium hydroxide would be required to completely neutralize 0.100 mole of hydrochloric acid?
The same molar amount, 0.100mol KOH. The reaction is in a 1:1 stoichiometric ratio: HCl + KOH --> KCl + H2O.
