Question

A 50.0 g sample of sodium phosphate reacts with aluminum chloride. The reaction produces 37.2 grams of aluminum phosphate and 53.4 g sodium chloride. How many grams of aluminum chloride were involved in the reaction?

Answer 1

To find the mass of aluminum chloride involved in the reaction between sodium phosphate and aluminum chloride, you need to use the concept of stoichiometry. First, calculate the moles of sodium chloride produced, then convert the moles of sodium chloride to moles of sodium phosphate. Next, convert the moles of sodium phosphate to moles of aluminum chloride, and finally convert the moles of aluminum chloride to grams using its molar mass.

Step-by-step explanation:

To find the mass of aluminum chloride involved in the reaction, we need to use the concept of stoichiometry. The balanced equation for the reaction is: 2 Na3PO4 + 3 AlCl3 → 6 NaCl + Al2(PO4)3. From the given information, we know that 50.0 g of sodium phosphate produced 53.4 g of sodium chloride. Using the stoichiometry of the balanced equation, we can calculate the number of moles of sodium phosphate and convert it to moles of aluminum chloride. Finally, we can convert the moles of aluminum chloride to grams using its molar mass.

1. Calculate the moles of sodium chloride produced: 53.4 g NaCl × (1 mol NaCl/58.44 g NaCl) = 0.914 mol NaCl
2. Convert the moles of sodium chloride to moles of sodium phosphate using the stoichiometry of the balanced equation: 0.914 mol NaCl × (2 mol Na3PO4/6 mol NaCl) = 0.305 mol Na3PO4
3. Convert the moles of sodium phosphate to moles of aluminum chloride using the stoichiometry of the balanced equation: 0.305 mol Na3PO4 × (3 mol AlCl3/2 mol Na3PO4) = 0.458 mol AlCl3
4. Finally, convert the moles of aluminum chloride to grams using its molar mass: 0.458 mol AlCl3 × (133.34 g AlCl3/1 mol AlCl3) = 61.0 g AlCl3