If 1 L HCl has a molarity of 0.10 M, what molar concentration of 1 L AgNO 3 is needed for HCl to completely react? How much product (moles and volume) will be formed?
HCl(aq)+ AgNO3(aq) ->AgC(s)+HNO3(aq)
typo in equation.
HCl + AgNO3 ==> AgCl + HNO3
mols HCl = M x L = 0.1 x 1 = 0.1
Convert mols HCl to mols AgNO3. 1 mol AgNO3 = 1 mol HCl; therefore, 0.1 mol HCl = 0.1 mol AgNO3.
Then M AgNO3 = mols AgNO3/L AgNO3. You know mols and L, solve for M.
To determine the molar concentration of AgNO3 needed for HCl to completely react, we can use the balanced equation and the stoichiometry of the reaction.
First, let's determine the stoichiometric ratio between HCl and AgNO3 from the balanced equation:
HCl(aq) + AgNO3(aq) -> AgCl(s) + HNO3(aq)
Looking at the balanced equation, we can see that the mole ratio between HCl and AgNO3 is 1:1. This means that for every 1 mole of HCl, we need 1 mole of AgNO3 to react completely.
Given that the molarity of the HCl solution is 0.10 M and the volume is 1 L, we can calculate the number of moles of HCl using the formula:
moles of solute = molarity × volume
moles of HCl = 0.10 M × 1 L = 0.10 moles of HCl
Since the stoichiometric ratio between HCl and AgNO3 is 1:1, we need 0.10 moles of AgNO3 to completely react with 0.10 moles of HCl.
To calculate the molar concentration of AgNO3 needed, we divide the moles of AgNO3 by the volume:
molarity of AgNO3 = moles of solute ÷ volume
molarity of AgNO3 = 0.10 moles ÷ 1 L = 0.10 M
Therefore, a molar concentration of 0.10 M AgNO3 is needed for HCl to completely react.
Now, let's determine the moles and volume of product formed, AgCl:
From the balanced equation, we can see that the stoichiometric ratio between AgCl and HCl is also 1:1.
Since the moles of HCl is 0.10 moles, the same number of moles of AgCl will be formed.
To calculate the volume of AgCl, we need to know the density or molar mass of AgCl and use the equation:
volume = moles ÷ molar mass or density
Without the density or molar mass information, we cannot calculate the exact volume of AgCl produced.