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first write balanced equation as follows : 2Na + S → Na2 S
45.3g Na & 105g Sulpher than Na2S = m=?
we know that
23g/mol na and 32.07g/mol sulpher gives 78.05 gm moles Na2S (23*2 +32.05 =78.05g/mol
moles Na available= 45.3 g/23 g/mole=1.96mole
1.96 * (1 mole of Na2S) / (2 mole of Na) = 0.98 mole of Na2S
(0.98 mol of Na2S) X 78.05 g/mol here moles get cancelled with each other hence we get
76.48 g of Na2S (Answer)
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Jun 30th, 2015
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