if 75.0 ml of 1.5m sodium hydroxide reacts with 100.0 ml of 2.0m phosphoric acid, to give sodium phosphate and water, how many grams of Na3Po4 may be produced?
First, I will assume you meant 1.5 M and 2.0 M. M stands for molarity, m stands for molality. Second, this is a limiting reagent problem and you know that because amounts are given for BOTH reactants.
3NaOH + H3PO4 ==> Na3PO4 + 3H2O
mols NaOH = M x L = ?
mols H3PO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols of the product.
Do the same and convert mols H3PO4 to mols of the product.
It is likely that these two values will not be the same; the correct value will ALWAYS be the smaller of the two and the reagent responsible for that will be the limiting reagent.
Using the smaller value for mols of the product, convert to grams. g Na3PO4 = mols Na3PO4 x molar mass Na3PO4 = ?.
To determine the number of grams of Na3PO4 that can be produced, we first need to calculate the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
Let's start by calculating the number of moles of both sodium hydroxide (NaOH) and phosphoric acid (H3PO4):
Moles of NaOH = Volume (in L) * Molarity
Moles of NaOH = 75.0 mL * (1 L / 1000 mL) * 1.5 mol/L
Moles of NaOH = 0.1125 mol
Moles of H3PO4 = Volume (in L) * Molarity
Moles of H3PO4 = 100.0 mL * (1 L / 1000 mL) * 2.0 mol/L
Moles of H3PO4 = 0.200 mol
Next, we need to determine the stoichiometric ratio between Na3PO4 and H3PO4. From the balanced chemical equation, we can see that 1 mole of H3PO4 reacts with 1 mole of Na3PO4.
Since there is an equal mole ratio, the limiting reactant will be the one with the fewer moles. In this case, NaOH has fewer moles (0.1125 mol) compared to H3PO4 (0.200 mol). Therefore, NaOH is the limiting reactant.
To find the number of moles of Na3PO4 that can be produced, we need to consider the stoichiometry. From the balanced chemical equation, we know that 1 mole of NaOH produces 1 mole of Na3PO4.
Moles of Na3PO4 = Moles of NaOH
Moles of Na3PO4 = 0.1125 mol
Finally, we can calculate the mass of Na3PO4 using its molar mass. The molar mass of Na3PO4 is 163.94 g/mol.
Mass of Na3PO4 = Moles of Na3PO4 * Molar Mass of Na3PO4
Mass of Na3PO4 = 0.1125 mol * 163.94 g/mol
Mass of Na3PO4 = 18.45 g
Therefore, approximately 18.45 grams of Na3PO4 may be produced.