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What mass of solid La(I03)3 (663.6 g/mol) is formed when 50.0 mL of 0.250 M La³+ are mixed

with 75.0 mL of 0.302 M I03?

asked
User Edmar
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1 Answer

6 votes

Answer:

To solve this problem, you need to use the mole ratios between the reactants to calculate how many moles of each reactant are used. The mole ratio between La³+ and La(I03)3 is 1:1 because there is one mole of La³+ for every mole of La(I03)3 that is produced. Similarly, the mole ratio between I03 and La(I03)3 is 3:1 because there are 3 moles of I03 for every mole of La(I03)3.

Using these mole ratios, you can calculate how many moles of each reactant are used in the reaction. First, you need to convert the volumes of the reactants to moles using their concentrations. The number of moles of La³+ is given by:

n(La³+) = (50.0 mL) * (0.250 M) = 12.5 moles

The number of moles of I03 is given by:

n(I03) = (75.0 mL) * (0.302 M) = 22.6 moles

Since the mole ratio between La³+ and La(I03)3 is 1:1, the number of moles of La(I03)3 produced is equal to the number of moles of La³+ used, which is 12.5 moles. The mass of La(I03)3 produced is then given by:

m(La(I03)3) = (12.5 moles) * (663.6 g/mol) = 8292.0 g

So, a mass of 8292.0 g of La(I03)3 is formed when 50.0 mL of 0.250 M La³+ are mixed with 75.0 mL of 0.302 M I03.

Step-by-step explanation:

answered
User Kevin Milner
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