For the first question:
= 1,312.5 g of Ba3(PO4)2
= 32.37 grams of HCl
Answer:
Explanation:
The balanced chemical equation for the reaction between potassium phosphate (K3PO4) and barium chloride (BaCl2) is:
2 K3PO4 + 3 BaCl2 → Ba3(PO4)2 + 6 KCl
This means that for every 2 moles of K3PO4 that react, we get 1 mole of Ba3(PO4)2 produced.
So, if we have 4.36 mol of K3PO4, we can calculate the amount of Ba3(PO4)2 produced as follows:
4.36 mol K3PO4 × (1 mol Ba3(PO4)2 / 2 mol K3PO4) = 2.18 mol Ba3(PO4)2
To convert this to grams, we need to use the molar mass of Ba3(PO4)2, which is:
3 × 137.33 g/mol (3 Ba atoms) + 2 × 94.97 g/mol (2 P atoms) + 8 × 16.00 g/mol (8 O atoms) = 601.93 g/mol
So, the mass of Ba3(PO4)2 produced is:
2.18 mol Ba3(PO4)2 × 601.93 g/mol = 1,312.5 g
Therefore, 4.36 mol of potassium phosphate would produce 1,312.5 grams of barium phosphate.
For the second question:
The balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl) is:
2 Al + 6 HCl → 2 AlCl3 + 3 H2
This means that for every 6 moles of HCl used, we get 2 moles of AlCl3 produced.
We are given that 39.5 g of AlCl3 is produced, so we can use its molar mass to calculate the number of moles produced:
39.5 g AlCl3 / 133.34 g/mol = 0.296 mol AlCl3
Since 2 moles of AlCl3 are produced for every 6 moles of HCl used, we can use this ratio to calculate the number of moles of HCl used:
0.296 mol AlCl3 × (6 mol HCl / 2 mol AlCl3) = 0.888 mol HCl
Finally, we can convert the number of moles of HCl used to grams using its molar mass:
0.888 mol HCl × 36.46 g/mol = 32.37 g HCl
Therefore, 39.5 g of AlCl3 would require 32.37 grams of HCl to react.