Question

What mass of agbr is formed when 57.8ml of 0.423m agno3 is treated with an excess of aqueous hydrobromic acid?

A) 0.091 g
B) 0.183 g
C) 0.366 g
D) 0.549 g

Answer 1

The mass of AgBr formed from 0.423 M AgNO3 solution and hydrobromic acid is calculated using stoichiometry, but the result does not match the provided options, indicating a potential error in calculations or the answer choices provided.

Step-by-step explanation:

To determine the mass of AgBr formed, the reaction between silver nitrate (AgNO3) and hydrobromic acid needs to be considered:

• AgNO3(aq) + HBr(aq) → AgBr(s) + HNO3(aq)

This is a stoichiometric reaction where 1 mole of AgNO3 reacts with 1 mole of HBr to form 1 mole of AgBr. Since silver nitrate is the limiting reactant, we can calculate the mass of AgBr produced by first finding the moles of AgNO3:

Molarity (M) = moles of solute / liters of solution

Moles of AgNO3 = Molarity × Volume (in liters) = 0.423 M × 0.0578 L = 0.024454 moles

Using the molar mass of AgBr (187.77 g/mol), we then calculate:

Mass of AgBr = moles × molar mass = 0.024454 moles × 187.77 g/mol = 4.594 g

However, none of the options A) 0.091 g, B) 0.183 g, C) 0.366 g, D) 0.549 g match this calculation. Since this seems to be an error, one should review the calculations or the given answer options again for accuracy.