Answer to Question #108364 in Chemistry for jill

Question

7. What mass of copper(II) sulfate would remain
after removing all the water from 57 g of
CuSO4 · 5 H2O?
Answer in units of g.
12. A chemist wants to extract the gold from 49.06 g of AuCl3 · 2 H2O (gold(III) chloride
dihydrate) by electrolysis of an aqueous solution. What mass of gold could be obtained
from this sample?
Answer in units of g.

Answer 1

Solution (7):

M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 9*Ar(O) + 10*Ar(H);

M(CuSO₄·5H₂O) = 63.546 + 32.076 + 9*15.9998 + 10*1.008 = 249.7 (g/mol).

M(CuSO₄) = Ar(Cu) + Ar(S) + 4*Ar(O) = 63.546 + 32.076 + 4*15.9998 = 159.62 (g/mol).

Let's find the percentage of copper(II) sulfate in CuSO₄·5H₂O:

w(CuSO₄) = M(CuSO₄) / M(CuSO₄·5H₂O) = (159.62) / (249.7) = 0.639247.

Then,

mass of copper(II) sulfate = m(CuSO₄) = w(CuSO₄) * m(CuSO₄·5H₂O);

m(CuSO₄) = 0.639247 * 57 = 36.44 g

m(CuSO₄) = 36.44 g.

Answer (7): 36.44 g of copper(II) sulfate (CuSO₄) would remain.

Solution (12):

Chemical reaction scheme:

AuCl₃ · 2H₂O --> Au

M(AuCl₃ · 2H₂O) = Ar(Au) + 3*Ar(Cl) + 4*Ar(H) + 2*Ar(O);

M(AuCl₃ · 2H₂O) = 196.9666 + 3*35.457 + 4*1.008 + 2*15.9998 = 339.3692 (g/mol).

M(Au) = 196.9666 (g/mol).

Let's find the percentage of gold (Au) in AuCl₃·2H₂O:

w(Au) = M(Au) / M(AuCl₃·2H₂O) = (196.9666) / (339.3692) = 0.58039.

Then,

mass of gold = m(Au) = w(Au) * m(AuCl₃·2H₂O);

m(Au) = 0.58039 * 49.06 = 28.4739 = 28.47 g

m(Au) = 28.47 g.

Answer (12): 28.47 g of gold (Au) could be obtained.