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How many moles of AgCl will be produced from 83.0 g of AgNO3 assuming NaCl is available in excess?
Wiki User
∙ 6y ago
X = 0.489 moles of AgCl produced
Wiki User
∙ 6y ago
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How many moles of silver chromate (Ag2CrO4) will be produced from 4 mol of silver nitrate (AgNO3)?
6
What is the balanced equation for silver nitrate and sodium hydroxide?
Assuming double displacement, AgNO3 + NaOH --> AgOH + NaNO3
How many grams of silver nitrate can be produced by the reaction of 100.0 ml of 0.20 M silver nitrate and 100.0 ml of 0.15 M calcium chloride?
2.9 g AgCl -Willy Willy Cheese Fried
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g?
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g then bromide=0.3220g and AgBr precipitate=0.6964.
What do you get from AgNO3 H2O?
AgNO3 + H2O ---> HNO3 + AgOH
4.02 grams of AgNO3 react with excess NaCl how many grams of silver nitrate are produced?
AgNo3(aq) + NaCl(aq) -------------->AgCl (ppt) + No3(-) + Na(+) no silver nitrates are produced it is all consumed. only silver chloride is produced and precipitate . free nitrate and free sodium ions are produced but do not react with each other 1 Mole AgNo3 ------->169.9 gm 1 Mole Nacl ------->58.4 gm 1 Mole AgCl ------->143.3 gm 4.02 gm AgNO3 = (4.02 / 169.9) = 0.02366 M AgCl produced = 0.02366 M = 3.39 gm
What mass of the precipitate could be produced by adding 100.0ml of 0.887 m agno3 to a na3po4 solution assume that the sodium phosphate reactant is present in excess?
12.4g
Calculate the mass of AgBr is formed when 35.5mL of 0.184 M AgNO3 is treated with an excess of aqueous hydrobromic acid HBr?
The mass of solid AgBr that is produced when 100.0 ml of 0.150 M AgNO3 is added to 20.0 ml of 1.00 M NaBr 2.82g.
10 grams of NaCl react with excess AgNO3. How many grams of solid AgCl are produced in this reaction?
trick question guys, there is none. this question should be deleted for such stupidity. FOOCK CHEMISTRY!
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 Ag are produced when starting with 6.2 moles of AgNO3?
6,2 moles of silver
How many moles of silver chromate (Ag2CrO4) will be produced from 4 mol of silver nitrate (AgNO3)?
6
What is the balanced equation for silver nitrate and sodium hydroxide?
Assuming double displacement, AgNO3 + NaOH --> AgOH + NaNO3
How much AgCl is produced when 4.22 grams of AgNO3 react with 7.73 grams of AlCl3?
3.56 g 3 AgNO3 + AlCl3 --> 3 AgCl + Al(NO3)3 AgCl3 is the limiting reagent (I checked), so: 4.22 g AgNO3 * (1 mol AgNO3/169.88 g AgNO3) * (3 mol AgNO3/3 mol AgCl3) * (143.32 mol AgCl3/1 mol AgCl3) =3.56 g AgCl3See the Related Questions to the left for more information about solving stoichiometry problems of this nature.
How many grams of silver nitrate can be produced by the reaction of 100.0 ml of 0.20 M silver nitrate and 100.0 ml of 0.15 M calcium chloride?
2.9 g AgCl -Willy Willy Cheese Fried
How many mol of AgCl can be precipitated by adding a solution containing 0.100 mol of AgNO3 to a solution containg excess NaCl?
Of course 0.100 mol of AgCl.
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g?
A sample of 0.3220 g of an ionic compound containing the bromide ion Br is dissolved in water and treated with an excess of AgNO3 If the mass of the AgBr precipitate that forms is 0.6964 g then bromide=0.3220g and AgBr precipitate=0.6964.
