Na2S2O3 + AgBr NaBr + Na3ſAg(S203)2] What is the mass of NaBr that will be produced from 42.7 g of AgBr?

Mass of NaBr produced  = 23.67 g

Explanation:

Given data:

Mass of AgBr = 42.7 g

Mass of NaBr produced = ?

Solution:

Chemical equation:

2Na₂S₂O₃ + AgBr    →    NaBr + Na₃(Ag(S₂O₃)₂

Number of moles of AgBr:

Number of moles = mass/molar mass

Number of moles = 42.7 g/ 187.7 g/mol

Number of moles = 0.23 mol

now we will compare the moles of AgBr with NaBr.

             AgBr        :         NaBr

                1            :           1

              0.23       :         0.23

Mass of NaBr:

Mass = number of moles × molar mass

Mass = 0.23 mol × 102.89 g/mol

Mass = 23.67 g

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l