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Ghella [55]
4 years ago
9

Please help me ASAP Will mark brainliest

Chemistry
1 answer:
Tomtit [17]4 years ago
5 0
The first one I believe
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A mixture 21.7 g NaCl 3.74 g kcl and 9.76 g licl how many moles of nacl are in this mixture
slava [35]
Moles (mol) = mass (g) / molar mass (g/mol)

Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>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 = <span>9.76 g
</span>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.
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