Lab calculations worksheet with solutions
1) Using the following equation:
2MnO4 + 5H2O2 + 6H --> 5O2 + 2Mn + 8H2O
If 36.44 mL of a 0.01625M KMNO4 solution are required to completely oxidize 25.00 mL of H2O2, calculate the molarity of H2O2
M (Base) = (Liters (Acid) x Molarity (Acid) x ratio of Base/Acid)/(Liters Base)
(.03644 x .01625 x 2.5) / .02500 = .05922M
2) A 100g sample of an unknown alkali metal sulfate is dissolved in water. When excess barium nitrate is added, 164.31g of insoluble barium sulfate is produced. What is the identity of the original sulfate compound?
Write the equation:
Ba(NO3) + X2SO4 --> BaSO4 + X2NO3 (x = alkali ion)
Since Sulfate has a 2- charge and the alkali is 1+ , you know there are 2 alkali ions.
Determine weights:
BaSO4 = 233.37 g/mol
SO4 = 96.02 g/mol
Calculate moles of BaSO4 used in the experiment:
163.41g / (233.37 g/mol )= .704 mol Ba & .704mol SO4
Calculate grams of sulfate:
.704mol SO4 x 96.02 g/mol = 67.63 g
Calculate grams of Alkali:
So out of the 100g used in the experiment only 32.40192g was the weight of the Alkali
100 – 67.63 = 32.37
Calculate moles of Alkali:
.704mol SO4 x ( 2 X) / (1 SO4 ) = 1.408 mol
Calculate weight of Alkali:
(32.37 g)/(1.408 mol) = 22.99 g/mol …..look at periodic table and you will find that is the weight of Sodium
3) 55.0 mL of .002M HCl was required to titrate a 30.0mL sample of Ba(OH)2 to its endpoint, what is the molarity of the barium hydroxide solution?
2HCl + Ba(OH)2 --> BaCl2 + 2H2O
M (Base) = M (Base) = (Liters (Acid) x Molarity (Acid) x ratio of Base/Acid) / (Liters Base)
(0.55L x .002M x .5) /.030L = 0.0018M
4) 60.0 ml of a solution containing calcium chloride is mixed with excess silver nitrate. If 0.23g of silver chloride is produced what was the molarity of the calcium chloride solution?
CaCl2 + 2 AgNO3 --> 2 AgCl + Ca(NO3)2
AgCl = 143.32 g/mol
(0.23g )/(143.32 g/mol) = .0016 mol AgCl x (1 CaCl2) / (2 AgNO3) = .000803 mol
(.000803 mol) / (.060 L) = .031 M
5) A metal, M3+, was converted to the sulfate M2SO4. Then a solution of the sulfate was treated with barium chloride to give barium sulfate crystals, which were filtered off. If 1.200 g of the metal gave 6.026 g of barium sulfate, what is the identity and atomic weight of the metal?
Write the equation:
M2(SO4)3 + 3BaCl2 --> 2MCl3 + 3BaSO4
Since Sulfate has a 2- charge and the alkali is 3+ , you know there are 3 alkali ions.
Determine weights:
BaSO4 = 233.37 g/mol
SO4 = 96.02 g/mol
Calculate moles of BaSO4 used in the experiment:
6.026g / (233.37g/mol) = .02582 mol Ba & .02582mol SO4
Calculate grams of sulfate:
.02582 mol SO4 x 96.02 g/mol = 2.479
Calculate moles of Metal:
.02585mol SO4 x ( 2 M ) / (3 SO4 ) = .0172 mol
Calculate weight of Metal:
(1.2 g) / (.0172 mol) = 69.76 g/mol …..look at periodic table and you will find that is the weight of Gallium