I mix 300.0ml of the barium nitrate solution from 1c with 100ml of a 1.50M lithium phosphate solution. The reaction produces solid barium phosphate
a. Write a balanced equation for the reaction.
b. Which is my limiting reagent?
c. How much barium phosphate may be produced in the reaction?
To answer these questions, we first need to write a balanced chemical equation for the reaction between barium nitrate and lithium phosphate. Then we can determine the limiting reagent and calculate the amount of barium phosphate produced.
a. The equation for the reaction between barium nitrate (Ba(NO3)2) and lithium phosphate (Li3PO4) can be written as follows:
Ba(NO3)2 + 3Li3PO4 -> 3LiNO3 + Ba3(PO4)2
This balanced equation shows that for every 1 mole of barium nitrate, we need 3 moles of lithium phosphate to react completely.
b. To determine the limiting reagent, we compare the stoichiometric ratios of the reactants to see which one is present in the limited amount. In this case, we have:
Barium nitrate: 300.0 ml = 0.3 L
Lithium phosphate: 100 ml = 0.1 L
Concentration of lithium phosphate: 1.50 M
To find the number of moles of each reactant, we use the formula:
moles = volume (in liters) x concentration (in moles per liter)
For barium nitrate:
moles of Ba(NO3)2 = 0.3 L x 1 C = 0.3 moles
For lithium phosphate:
moles of Li₃PO₄ = 0.1 L x 1.50 M = 0.15 moles
Now, looking at the balanced equation, we see that we need 3 moles of Li₃PO₄ for every 1 mole of Ba(NO3)₂. Since we have fewer moles of Ba(NO3)₂ (0.3 moles) compared to the stoichiometric ratio, it is the limiting reagent.
c. Now that we know that barium nitrate is the limiting reagent, we use its stoichiometry to calculate the amount of barium phosphate produced. From the balanced equation, we see that 1 mole of Ba(NO3)₂ produces 1 mole of Ba₃(PO₄)₂.
Therefore, the moles of Ba₃(PO₄)₂ produced will be equal to the moles of Ba(NO3)₂, which is 0.3 moles. To find the mass of Ba₃(PO₄)₂ produced, we need to multiply the moles by the molar mass of Ba₃(PO₄)₂.
Molar mass of Ba₃(PO₄)₂ = 1 * (3 * atomic mass of Ba) + 2 * (atomic mass of P + 4 * atomic mass of O)
Now, by substituting the values and calculations into the formula, you can determine the mass of barium phosphate produced.