Stoichiometric Calculations: Write the complete solutions:
What mass of solid La(IO3)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 IO3- ?
First, we need to determine the limiting reactant between La3+ and IO3- in order to find the theoretical yield of La(IO3)3.
Step 1: Write the balanced chemical equation for the reaction between La3+ and IO3-:
2 La3+ + 3 IO3- → La(IO3)3
Step 2: Calculate the moles of La3+ and IO3- in each solution:
Moles of La3+ = (50.0 ml) x (0.250 mol/L) = 0.0125 mol
Moles of IO3- = (75.0 ml) x (0.302 mol/L) = 0.02265 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, we need to compare the ratio of moles of La3+ and IO3- to the stoichiometry of the balanced chemical equation.
From the balanced chemical equation, we see that 2 moles of La3+ react with 3 moles of IO3-.
Therefore, La3+ is the limiting reactant because it will be completely consumed before all of the IO3- is used up.
Step 4: Calculate the moles of La(IO3)3 formed using the limiting reactant:
Since La3+ is the limiting reactant, we can use the moles of La3+ to calculate the moles of La(IO3)3 formed.
From the balanced chemical equation, we see that it forms in a 1:1 molar ratio with La3+.
Therefore, moles of La(IO3)3 = moles of La3+ = 0.0125 mol
Step 5: Calculate the mass of La(IO3)3 formed:
Mass of La(IO3)3 = moles of La(IO3)3 x molar mass of La(IO3)3
Mass of La(IO3)3 = 0.0125 mol x 663.6 g/mol = 8.295 g
Therefore, when 50.0 ml of 0.250 M La3+ are mixed with 75.0 ml of 0.302 M IO3-, the mass of solid La(IO3)3 formed is 8.295 grams.