Answer to Question #15733 in Other Chemistry for Porscha
Question #15733
How many grams of water are contained in 2.78g og CaSo4*2H2O
Expert's answer
Molecular mass of CaSO4*2H2O is 172 g/mol. The one mole of CaSO4*2H2O contains
two moles of water. The amount of CaSO4*2H2O is
n(CaSO4*2H2O)=m(CaSO4*2H2O)/M(CaSO4*2H2O)=2.78 g / 172 g/mol = 0.0162 mol.
n(H2O)=2*n(CaSO4*2H2O)=0.0323 mol. m(H2O)=n(H2O)*M(H2O)=0.0323*18=0.582 g.
two moles of water. The amount of CaSO4*2H2O is
n(CaSO4*2H2O)=m(CaSO4*2H2O)/M(CaSO4*2H2O)=2.78 g / 172 g/mol = 0.0162 mol.
n(H2O)=2*n(CaSO4*2H2O)=0.0323 mol. m(H2O)=n(H2O)*M(H2O)=0.0323*18=0.582 g.
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Everything works, thank you (Operator and Expert) so much! I will bookmark this page for future reference. Excellent job on the comments to by the way, now I can study and understand what is happening for the future!
#198641 on Mar 2018
#198641 on Mar 2018 
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