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How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Wiki User
∙ 13y ago
Balanced equation first!
AgNO3 + NaCl -> AgCl + NaNO3
all one to one, get moles AgNO3
3.82 moles NaCl (1 mole AgNO3/1 mole NaCl)
= 3.82 moles AgNO3
-------------------------------
Molarity = moles of solute/Liters of solution
0.117 M AgNO3 = 3.82 moles AgNO3/Liters
Liters = 3.82/0.117
= 32.6 Liters
which is
32600 milliliters
which is unreasonable; check answer if you can
Wiki User
∙ 13y ago
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How many moles of agno3 are needed to prepare 0.50 l of a 4.0 m solution?
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
What is the molarity of a solution if 255 grams AgNO3 is dissolved in 1500 mL of solution?
Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------
How many moles of AgNO3 does 85 grams of AgNO3 represents?
85 grams of AgNO3 represents 0,.5 moles.
How many moles of potassium hydroxide are required to make 200cm3 of 0.5M solution?
0.1 moles
How many moles of silver are present in 32.46g of AgNO3?
AgNO3 is 169.87 g/mol. Ag is 107.87 g/mol meaning that Ag is 63.5% of AgNO3. 63.5% of 32.46g is 20.61g. 20.61g / 107.87 g/mol = .191 moles.
How many moles of agno3 are needed to prepare 0.50 l of a 4.0 m solution?
Molarity = moles of solute/liters of solution or, for our purposes moles of solute = liters of solution * Molarity moles of AgNO3 = 0,50 liters * 4.0 M = 2.0 moles of AgNO3 needed --------------------------------------
What is the molarity of a solution if 255 grams AgNO3 is dissolved in 1500 mL of solution?
Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------
How many moles of AgNO3 does 85 grams of AgNO3 represents?
85 grams of AgNO3 represents 0,.5 moles.
How many grams of magnesium hydroxide required to make 1M of the solution?
the formula is no. moles is mass / molecular mass. As the number of moles is 1, the mass required will be exactly the same as the molecular mass, which is 58.32g
How many moles are in 680g of AgNO3?
Roughly 4 moles.
How many moles of Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
How many moles of potassium hydroxide are required to make 200cm3 of 0.5M solution?
0.1 moles
How many moles of silver are present in 32.46g of AgNO3?
AgNO3 is 169.87 g/mol. Ag is 107.87 g/mol meaning that Ag is 63.5% of AgNO3. 63.5% of 32.46g is 20.61g. 20.61g / 107.87 g/mol = .191 moles.
How many moles of silver of ions are presented in 32.46 g of AgNO3?
The number of moles is 0,19.
How much distilled water should be used to dissolve 3.33 g of silver nitrate?
It depends on the strength of the solution you want. 3.33g of AgNO3 is 0.0196 moles. Ish. So from there, you need to work out how much water is needed to create your required strength. 1.42 moles of silver nitrate is the maximum you can dissolve in 100g of water.
What mass of solid agcl is obtained when 25 ml of 0.068m agno3 reacts with excess of aqueous hcl?
By definition, No. of moles = given mass/molecular mass; and also by definition, molar concentration of a solute means the number of moles of solute per liter of solution. Therefore, 25 mL of 0.068 M AgNO3 contains AgNO3 = (0.068 * 25) / 1000 = 0.0017 mol of AgNO3.The equation for the reaction is AgNO3 + HCl -> AgCl + HNO3, showing that 0.0017 mol of AgNO3 gives 0.0017 mol of AgCl. The molecular mass of AgCl = 107+35.5 = 143.5 gTherefore, the mass of AgCl produced by the reacion = No. of moles*molecular mass = .0017*143.5 = 0.24g, to the justified number of significant digits.100 mL of 0.068 M AgNO3 contains AgNO3 = 0.068 molSo,Therefore,
How many moles of NaCl is required to make 1L of a 140mM solution?
2g
