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Question

A student needed magnesium sulfate heptahydrate, but could not read what the hydrate of magnesium sulfate was in the bottle. The student decided to determine experimentally that the substance was in fact magnesium Sulfate heptahydrate. A student weighs out 10.000 grams of MgSO4 from the bottle. After heating, the sample mass was 5.720 grams. The chemical equation that occurs upon heating a hydrate is :
MgSO4 X 7H2O (s)------->    MgSO4 (s) + 7 H2O (g)

a. In theory how much water should the reaction produce if the solid in the bottle is MgSO4 X7H2O (s)?

B. How much water was actually removed from heating
C. How many mole of water was removed?
D. How many Mole of magnesium sulfate was produced?
E. What is the experimentally determined formula for the hydrate in the bottle?
F. Was the sample MgSo4 X 7H2O?

Quick response Plz

matthewthegenuis · Accepted Answer

A Divide the mass MgSO4 X7H2O that reacted by the Mr to calcualte moles (10g/239.4) = 0.042 moles. Moles water produced is 0.042 moles x 7 (as equation is 1:7 MgSO4 X7H2O: H2O) = 0.29 moles. Mass water = 0.29 moles x 18 (Mr of H2O) = 5.26g produced in theory.
B. Amount actually produced = 10g - 5.720g = 4.28g.
C. Moles H2O produced - 4.28g/18 (Mr of water) = 0.24 moles.
D. Moles MgSO4 produced is 5.72g/120.4 (Mr MgSO4) = 0.0475 moles.
E. Mr MgSO4xH2O = 10g/0.0475 moles (equation 1:1 MgSO4:MgSO4x7H2O) = 210.5. Mr MgSO4 = 120.4. 210.5 - 120.4 = 90.13/18 (Mr of water) = 5. Therefore the experimental formula is MgSO4.5H2O
F. The sample was therefore not MgSO4.7H2O.

Anonymous · Answer

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Anonymous · Answer

the change in mass will be the amount of water that was originally in the reactant.

ananthi · Answer

a)4.28g
b)
c)0.24 mole
d)0.0475 mole

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HELP HURRY WHAT is the answer?

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