Determine the percentages of Na and K in a 0.6128 g sample containing NaCl and KCl treated with AgNO3 yielded 1.039 g of dried AgCl (molecular weight = 143.32g/mol). NaCl and KCl have molecular weights of 58.44g/mol and 74.55g/mol, respectively
To determine the percentages of Na and K in the sample, we first need to calculate the amount of NaCl and KCl in the sample.
1. Calculate the moles of AgCl produced:
moles of AgCl = mass of AgCl / molecular weight of AgCl
moles of AgCl = 1.039 g / 143.32 g/mol = 0.00725 mol AgCl
2. Since AgCl is produced from 1:1 molar ratio of AgCl to Cl in NaCl and KCl:
moles of NaCl or KCl = moles of AgCl
moles of NaCl or KCl = 0.00725 mol
3. Calculate the mass of NaCl and KCl in the sample:
mass of NaCl = moles of NaCl x molecular weight of NaCl
mass of NaCl = 0.00725 mol x 58.44 g/mol = 0.4242 g NaCl
mass of KCl = moles of KCl x molecular weight of KCl
mass of KCl = 0.00725 mol x 74.55 g/mol = 0.5408 g KCl
4. Calculate the percentages of Na and K in the sample:
% Na = (mass of NaCl / total mass of sample) x 100%
% Na = (0.4242 g / 0.6128 g) x 100% = 69.27%
% K = (mass of KCl / total mass of sample) x 100%
% K = (0.5408 g / 0.6128 g) x 100% = 88.37%
Therefore, the sample contains approximately 69.27% Na and 88.37% K.