What mass of AgCl will be obtained by adding an excess of HCl to a solution of 0.34 g of AgNO3 ?
AgNO3 + HCl ==> AgCl + HNO3
mols AgNO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols AgNO3 to mols AgCl.
Now convert mols AgCl to grams. g = mols x molar mass.
I want the answer
To determine the mass of AgCl obtained, you need to understand the reaction between AgNO3 (silver nitrate) and HCl (hydrochloric acid). The balanced chemical equation for this reaction is:
AgNO3 + HCl → AgCl + HNO3
From the balanced equation, you can see that 1 mole of AgNO3 reacts with 1 mole of HCl to produce 1 mole of AgCl.
First, you need to find the number of moles of AgNO3 in the given solution:
Molar mass of AgNO3 = atomic mass of Ag + atomic mass of N + 3 * atomic mass of O
= (107.87 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol)
= 169.87 g/mol
Number of moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
= 0.34 g / 169.87 g/mol
Next, you can calculate the number of moles of AgCl formed using the mole ratio from the balanced equation:
Number of moles of AgCl = Number of moles of AgNO3
Finally, you can determine the mass of AgCl formed using the molar mass of AgCl:
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Mass of AgCl = Number of moles of AgCl * molar mass of AgCl
By plugging in the calculated moles, you will get the mass of AgCl obtained in grams.