Question

100 ml of 2.2 normal HCL is added to underdevelopment to in AG and O3 solution is the molality of nitrate ions in the resulting mixture will be​

Answer 1

Explanation:

You're dealing with a double replacement reaction in which two soluble compounds react in aqueous solution to form an insoluble solidthat precipitates out of solution.

Silver nitrate, AgNO3, a soluble ionic compound, will react with hydrochloric acid, HCl, a strong acid that dissociates completely in aqueous solution, to form the insoluble silver chloride, AgCl,

AgNO3(aq]+HCl(aq]→AgCl(s]⏐↓+HNO3(aq]

The net ionic equation, which you get after removing the spectator ions, looks like this

Ag+(aq]+Cl−(aq]→AgCl(s]⏐⏐↓

Notice that the reactants take part in the reaction in a 1:1 mole ratio, which means that the reaction will consume equal numbers of molesof silver nitrate and hydrochloric acid, which produce silver cations, Ag+, and chloride anions, Cl−, respectively.

Your goal now is to figure out how many moles of each reactant you're mixing together.

For starters, use silver nitrate's molar mass to determine how many moles you get in that 1.40-g sample of silver nitrate

1.40g⋅1 mole AgNO3169.87g=0.008242 moles AgNO3

Now use the molarity and volume of the hydrochloric acid solution to determine how many moles of acid you have

c=nV⇒n=c⋅V

nHCl=1.50 M⋅5.00⋅10−3L=0.00750 moles HCl

Since you have fewer moles of chloride anions,

Answer 2

Answer:

Silver nitrate, AgNO3, a soluble ionic compound, will react with hydrochloric acid, HCl, a strong acid that dissociates completely in aqueous solution, to form the insoluble silver chloride, AgCl,

AgNO3(aq]+HCl(aq]→AgCl(s]⏐↓+HNO3(aq]

The net ionic equation, which you get after removing the spectator ions, looks like this

Ag+(aq]+Cl−(aq]→AgCl(s]⏐⏐↓

Notice that the reactants take part in the reaction in a 1:1 mole ratio, which means that the reaction will consume equal numbers of molesof silver nitrate and hydrochloric acid, which produce silver cations, Ag+, and chloride anions, Cl−, respectively.

Your goal now is to figure out how many moles of each reactant you're mixing together.

For starters, use silver nitrate's molar mass to determine how many moles you get in that 1.40-g sample of silver nitrate

1.40g⋅1 mole AgNO3169.87g=0.008242 moles AgNO3

Now use the molarity and volume of the hydrochloric acid solution to determine how many moles of acid you have

c=nV⇒n=c⋅V

nHCl=1.50 M⋅5.00⋅10−3L=0.00750 moles HCl

Since you have fewer moles of chloride anions,