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Mixing Solutions

Mixing Solutions. Putting it together. Two solutions are combined. How many moles of the products will be formed? How many grams of product will be formed? . To calculate grams, we need to know moles. If we have Molarity and Litres, we can find the number of moles. M = moles

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Mixing Solutions

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  1. Mixing Solutions Putting it together

  2. Two solutions are combined. How many moles of the products will be formed? How many grams of product will be formed? • To calculate grams, we need to know moles. • If we have Molarity and Litres, we can find the number of moles. • M = moles • litres • Then, it is a limiting reagent problem. • Check mole ratios to see which reactant is limited. • Convert moles to grams.

  3. Two solutions are combined. One is 200.0cm3 of 0.800M BaC2O4, and the other is 350.0cm3 of 0.400M Fe2(SO4)3 . How many moles of iron (III) oxalate will be formed? How many grams of barium sulfate will be formed? • 3BaC2O4 + Fe2(SO4)3 3BaSO4 + Fe2(C2O4)3 M = moles litres 0.800 = moles 0.200 moles = 0.160 mol BaC2O4 0.400 = moles 0.350 moles = 0.140 mol Fe2(SO4)3

  4. We have: 0.160 mol BaC2O4 • 0.140 mol Fe2(SO4)3 • 3BaC2O4 + Fe2(SO4)3 3BaSO4 + Fe2(C2O4)3 • Since we only have 0.160 mol of BaC2O4it is the limiting reagent. Therefore we use the moles of BaC2O4 to do the calculations.

  5. 3BaC2O4 + Fe2(SO4)3 3BaSO4 + Fe2(C2O4)3 • Now we can use the limiting number of moles to calculate grams. • 233.3gx 0.160 mol = 37.3g • 1 mole BaSO4 • 37.3g BaSO4 formed • 0.053 mol Fe2(C2O4)3 formed

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