Homework Answers
Given
weight of hydrated material = 8.55 g
Weight after losing hydrate = 7.55 g (only Na2CO3)
Molar mass of Na2CO3 = 106 g/mol
No. of moles of Na2CO3 = 7.55 g / 106 g/mol = 0.0712 moles
same no. of moles of Na2CO3 with xH2O will be present in intial material
so Molar mass of initial material = 8.55 g / 0.0712 moles = 120 g/mol
Molar mass of Na2CO3.xH2O = 120 g/mol
Molar mass of xH2O = 120 - 106 = 18 g/mol
which is molar mass of 1 H2O so
molar ratio is 1:1 answer 1:1
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Calculate the theoretical yield and percent yield for
experiments 7, 8, and 9. For experiment 9, determine the
theoretical yield based on starting with 0.50g of 3-nitrobenzoic
acid. Show work
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g
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g
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Calculate the theoretical yield and percent yield for
experiments 7, 8, and 9. For experiment 9, determine the
theoretical yield based on starting with 0.50g of 3-nitrobenzoic
acid. Show work
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g
Acid II (experiment 8): Weight of 3-nitrobenzoic acid = 1.75
g
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-1.75 mL concentrated HNO3 (16M)
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