Amixture 21.7 g nacl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture

Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NaCl = 21.7 g
Molar mass of NaCl = 58.4 g/mol
Hence, moles of NaCl = 21.7 g / 58.4 g/mol = 0.372 mol

Hence moles of NaCl in the mixture is 0.372 mol.

Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.

mole % = (moles of desired substance / Total moles of the mixture) x 100%

Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%

Total moles of mixture = moles of NaCl + KCl + LiCl

Mass of KCl = 3.74 g 
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g  / 74.6 g/mol = 0.050 mol

Mass of NaCl = 9.76 g
Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol

Total moles =  0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol

mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
                           = (0.372 mol / 0.652 mol) x 100%
                           = 57.06% 

Hence, mixture has 57.06% of NaCl as the mole percentage.

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.

The formula is (nh4)2co3