Find the theoretical yield of hydrated metal oxalate product 3. Balance the molecular equation that shows your metal salt, MNSO4.H20, reacting with oxalic acid (H2C204) to form MnC204.3H20 (enter a 1 if a stoichiometric coefficient is one).1 MnSO4.H20 +1 H2C204+2 H20 =1 MnC204.3H20+1 H2SO4correct, 1/14. Pyrolysis of MnC204.3H20 will follow one of the following molecular equations. Balance each molecular equation,2|MnC204.3H2O +102 =2 MnO6 H20 +4co,4 CO2a.b.3 MnC204.3H2O +202=1 Mn304 +9 H20 +6 CO21 MnC204.3H2O1 MNCO3 +3 H20 +1co1 COС.(a)correct, 1/1(b)correct, 1/1(c)correct, 1/15. For pyrolysis reaction (4a) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant0.36 gcorrect 1/16. For pyrolysis reaction (4b) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant0.39 gcorrect 1/17. For pyrolysis reaction (4c) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant0.58 gcorrect 1/1 DataResultMass of hydrated metal salt used4.000 gVolume of oxalic acid used (assume its concentration is, 4.000 g oxalic acid/50.00 mL soln)50.00 mLMass of oxalic acid used4.000 gMass of weigh dish + contents (after reaction)5.320 gMass of empty weigh dish1.881 gMass of filter paper0.101 gActual yield of hydrated metal oxalate ppt(A)3.862 gLimiting reactant(B) metal salt vTheoretical yield of hydrated metal oxalate product(C)Percent yield of hydrated metal oxalte product(D)%

#Calculation for theoretical yield: 169.017 g/mol Here the limiting reactant is the hydrated metal…

Data Result Mass of hydrated metal salt used 4.000 g Volume of oxalic acid used (assume its concentration is, 4.000 g oxalic acid/50.00 mL soln) 50.00 mL Mass of oxalic acid used 4.000 g Mass of weigh dish + contents (after reaction) 5.320 g Mass of empty weigh dish 1.881 g Mass of filter paper 0.101 g Actual yield of hydrated metal oxalate ppt (A) 3.862 g Limiting reactant (B) metal salt ♥ Theoretical yield of hydrated metal oxalate product (C) g Percent yield of hydrated metal oxalte product (D)

Data Result Mass of hydrated metal salt used 4.000 g Volume of oxalic acid used (assume its concentration is, 4.000 g oxalic acid/50.00 mL soln) 50.00 mL Mass of oxalic acid used 4.000 g Mass of weigh dish + contents (after reaction) 5.320 g Mass of empty weigh dish 1.881 g Mass of filter paper 0.101 g Actual yield of hydrated metal oxalate ppt (A) 3.862 g Limiting reactant (B) metal salt ♥ Theoretical yield of hydrated metal oxalate product (C) g Percent yield of hydrated metal oxalte product (D)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter4: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 138CP: Polychlorinated biphenyls (PCBs) have been used extensively as dielectric materials in electrical...

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Question

Find the theoretical yield of hydrated metal oxalate product

3. Balance the molecular equation that shows your metal salt, MNSO4.H20, reacting with oxalic acid (H2C204) to form MnC204.3H20 (enter a 1 if a stoichiometric coefficient is one).
1 MnSO4.H20 +
1 H2C204+
2 H20 =
1 MnC204.3H20+
1 H2SO4
correct, 1/1
4. Pyrolysis of MnC204.3H20 will follow one of the following molecular equations. Balance each molecular equation,
2|MnC204.3H2O +
102 =
2 MnO
6 H20 +
4co,
4 CO2
a.
b.
3 MnC204.3H2O +
202=
1 Mn304 +
9 H20 +
6 CO2
1 MnC204.3H2O
1 MNCO3 +
3 H20 +
1co
1 CO
С.
(a)correct, 1/1
(b)correct, 1/1
(c)correct, 1/1
5. For pyrolysis reaction (4a) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant
0.36 g
correct 1/1
6. For pyrolysis reaction (4b) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant
0.39 g
correct 1/1
7. For pyrolysis reaction (4c) (above), calculate the theoretical yield (in grams) of the solid product, if you use 1.0 g MnC204.3H20 and that oxygen is the excess reactant
0.58 g
correct 1/1

Data
Result
Mass of hydrated metal salt used
4.000 g
Volume of oxalic acid used (assume its concentration is, 4.000 g oxalic acid/50.00 mL soln)
50.00 mL
Mass of oxalic acid used
4.000 g
Mass of weigh dish + contents (after reaction)
5.320 g
Mass of empty weigh dish
1.881 g
Mass of filter paper
0.101 g
Actual yield of hydrated metal oxalate ppt
(A)
3.862 g
Limiting reactant
(B) metal salt v
Theoretical yield of hydrated metal oxalate product
(C)
Percent yield of hydrated metal oxalte product
(D)
%

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