You are assigned the following amount of CoCl2.
Volume of CoCl2 assigned and used 4.84 mL
Concentration of CoCl2 1.39 M
To this is added...
Mass of charcoal 0.23 g
Volume of 15.0 M ammonia 5.3 mL
Mass of ammonium chloride 1.38 g
The ammonium chloride was dissolved. The mixture was cooled. To this was added:
Volume of 10% (by weight) hydrogen peroxide 4.9 mL
(density = 1.00 g/mL)
Calculate:
Millimoles of CoCl2 used ____6.7276___________ mmol
b) Millimoles of NH3 theoretically required _______________ mmol
c) Millimoles of NH3 actually added _______________ mmol
d) Millimoles of NH4Cl theoretically required _______________ mmol
e) Millimoles of NH4Cl actually added _______________ mmol
f) Millimoles of H2O2 theoretically required _______________ mmol
g) Millimoles of H2O2 actually added:14.41mml
To calculate the millimoles of CoCl2 used, we need to use the given volume and concentration.
Given:
Volume of CoCl2 assigned and used: 4.84 mL
Concentration of CoCl2: 1.39 M
To calculate millimoles, we can use the formula:
Millimoles = Volume (in mL) x Concentration (in M)
Millimoles of CoCl2 used = 4.84 mL x 1.39 M = 6.7276 mmol
a) Millimoles of CoCl2 used: 6.7276 mmol
To calculate the millimoles of NH3 theoretically required, we need to know the stoichiometry of the reaction between CoCl2 and NH3. Since you haven't provided that information, it's not possible to calculate this value.
b) Millimoles of NH3 theoretically required: Not calculable without stoichiometry information
To calculate the millimoles of NH3 actually added, we need to use the given volume and concentration.
Given:
Volume of 15.0 M ammonia: 5.3 mL
Millimoles of NH3 actually added = 5.3 mL x 15.0 M = 79.5 mmol
c) Millimoles of NH3 actually added: 79.5 mmol
To calculate the millimoles of NH4Cl theoretically required, we need to know the stoichiometry of the reaction involving NH4Cl. Without that information, it's not possible to calculate this value.
d) Millimoles of NH4Cl theoretically required: Not calculable without stoichiometry information
To calculate the millimoles of NH4Cl actually added, we need to use the given mass.
Given:
Mass of ammonium chloride: 1.38 g
To calculate the millimoles of NH4Cl, you first need to convert the mass to moles by dividing by the molar mass.
Molar mass of NH4Cl = 14.01 g/mol (N) + 4(1.01 g/mol) (H) + 35.45 g/mol (Cl) = 53.49 g/mol
Millimoles of NH4Cl actually added = (1.38 g / 53.49 g/mol) x 1000 = 25.81 mmol
e) Millimoles of NH4Cl actually added: 25.81 mmol
To calculate the millimoles of H2O2 theoretically required, we again need more information about the stoichiometry of the reaction involving H2O2. Without that information, it's not possible to calculate this value.
f) Millimoles of H2O2 theoretically required: Not calculable without stoichiometry information
To calculate the millimoles of H2O2 actually added, we need to use the given volume.
Given:
Volume of 10% (by weight) hydrogen peroxide: 4.9 mL (density = 1.00 g/mL)
To calculate the millimoles of H2O2, you first need to convert the volume to mass by multiplying it by the density.
Mass of 10% H2O2 = 4.9 mL x 1.00 g/mL = 4.9 g
Divide the mass of H2O2 by its molar mass to obtain the millimoles.
Molar mass of H2O2 = 2(1.01 g/mol) (H) + 16.00 g/mol (O) = 34.02 g/mol
Millimoles of H2O2 actually added = (4.9 g / 34.02 g/mol) x 1000 = 143.95 mmol
g) Millimoles of H2O2 actually added: 143.95 mmol