Experiment 3-Pre Lab For desk numbers 3, 6, 9, 12, 15, 18, 21, and 24 Answer the

Question

Experiment 3-Pre Lab For desk numbers 3, 6, 9, 12, 15, 18, 21, and 24 Answer the following questions. Show all work and calculations to support your answers. Include the correct units and the proper number of significant figures on numerical answers. Circle your final 1. For the metal salt indicated in the procedure, you should use MnsO, H,O. Calculate the molar mass of MnSo, H,O. 2. Your oxalate synthesis product will be MnC,0,.3H,O. Calculate the molar mass of MnC,0,.3H,O. 3. Write a balanced molecular equation that shows your metal salt, MnSO H,O, reacting with oxalic acid (H,C,O,) and water to form MnGo,.3H,O. What will be the other product(s) of the reaction? Experiment 3 0xalate Stoichiometry 23

Explanation / Answer

1) We will need to know the atomic masses of the elements forming the compound.

Mn: 54.938 g/mol

S: 32.06 g/mol

O: 15.999 g/mol

H: 1.008 g/mol.

Molar mass of MnSO4.H2O = (1*54.938 + 1*32.06 + 4*15.999 + 2*1.008 + 1*15.999) g/mol = 169.009 g/mol.

2) We shall need the atomic mass of C.

C: 12.011 g/mol.

Molar mass of MnC2O4.3H2O = (1*54.938 + 2*12.011 + 4*15.999 + 6*1.008 + 3*15.999) g/mol = 196.999 g/mol.

3) The balanced chemical equation for the reaction is

MnSO4.H2O + H2C2O4 + 2 H2O ----------> MnC2O4.3H2O + H2SO4

H2SO4 is the other product of the reaction.

4a) Start by balancing the number of C atoms on both sides.

MnC2O4.3H2O (s) + O2 (g) ---------> MnO (s) + H2O (g) + 2 CO2 (g)

Next balance the number of H atoms on both sides.

MnC2O4.3H2O (s) + O2 (g) ---------> MnO (s) + 3 H2O (g) + 2 CO2 (g)

Finally balance the number of O atoms on both sides.

MnC2O4.3H2O (s) + O2 (g) ---------> MnO (s) + 3 H2O (g) + 2 CO2 (g)

Note that we have two extra O atoms on the left and one on the right; multiply the right by 2 and MnC2O4.3H2O by 2 to balance. The balanced chemical equation is

2 MnC2O4.3H2O (s) + O2 (g) ---------> 2 MnO (s) + 6 H2O (g) + 4 CO2 (g)

b) Start by balancing Mn atoms on both sides.

3 MnC2O4.3H2O (s) + O2 (g) ---------> Mn3O4 (s) + H2O (g) + CO2 (g)

Next balance the number of C atoms on both sides.

3 MnC2O4.3H2O (s) + O2 (g) ---------> Mn3O4 (s) + H2O (g) + 6 CO2 (g)

Next, we shall balance the number of H atoms.

3 MnC2O4.3H2O (s) + O2 (g) ---------> Mn3O4 (s) + 9 H2O (g) + 6 CO2 (g)

Finally balance the number of O atoms by adding 2 O2 on the left.

3 MnC2O4.3H2O (s) + 2 O2 (g) ---------> Mn3O4 (s) + 9 H2O (g) + 6 CO2 (g)

c) Start by balancing the number of C atoms on both sides.

MnC2O4.3H2O (s) ---------> MnCO3 (s) + H2O (g) + CO (g)

Next balance the number of H atoms.

MnC2O4.3H2O (s) ---------> MnCO3 (s) + 3 H2O (g) + CO (g)

The number of O atoms are balanced; hence, the reaction is balanced.

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