Answer:
400.87g of barium phosphate and 32.4g of HCL
Step-by-step explanation:
The balanced chemical equation for the reaction between potassium phosphate and barium nitrate is:
3 K3PO4 + 4 Ba(NO3)2 → 12 KNO3 + Ba3(PO4)2
According to the stoichiometry of the equation, for every 3 moles of potassium phosphate, 1 mole of barium phosphate is produced. Therefore:
1 mol Ba3(PO4)2 = 3 mol K3PO4
To convert the given quantity of potassium phosphate to moles, we can use its molar mass:
4.36 mol K3PO4 = 4.36 mol × 212.27 g/mol = 925.5912 g
Now we can use the stoichiometry to calculate the amount of barium phosphate produced:
1 mol Ba3(PO4)2 = 3 mol K3PO4
1 mol Ba3(PO4)2 = 3/4 mol Ba(NO3)2 (from the balanced equation)
Therefore, the amount of barium phosphate produced is:
4.36 mol K3PO4 × 1 mol Ba3(PO4)2 / 3 mol K3PO4 × 4 mol Ba(NO3)2 / 3 mol Ba3(PO4)2 × 601.93 g/mol Ba3(PO4)2 = 400.87 g
Therefore, 400.87 grams of barium phosphate are produced.
We need to know the balanced chemical equation for the reaction in order to determine the stoichiometry of the reactants and products. Let's assume that the reaction is:
2 Al + 6 HCl → 2 AlCl3 + 3 H2
This equation tells us that 6 moles of HCl are required to produce 2 moles of AlCl3. The molar mass of AlCl3 is:
1 Al atom × 26.98 g/mol + 3 Cl atoms × 35.45 g/mol = 133.34 g/mol
Therefore, 39.5 g of AlCl3 represents:
39.5 g ÷ 133.34 g/mol = 0.296 moles of AlCl3
Since the reaction produces 2 moles of AlCl3 for every 6 moles of HCl, we can use a ratio to find the number of moles of HCl required:
0.296 moles AlCl3 × (6 moles HCl / 2 moles AlCl3) = 0.888 moles HCl
Finally, we can convert the number of moles of HCl to grams:
0.888 moles HCl × 36.46 g/mol = 32.4 g HCl
Therefore, 32.4 g of HCl was used in the reaction.