Question

Calculate the maximum grams of product for the reaction described below by constructing a BCA table and determining the maximum grams of possible product. Complete Parts 1-2 before submitting your answer. Ca

3
​
(PO
4
​
)
2
​
( s)+3H
2
​
SO
4
​
(aq)→3CaSO
4
​
( s)+2H
3
​
PO
4
​
(aq) reaction occurs starting with 1.00 kg of Ca
3
​
(PO
4
​
)
2
​
and 1.00 kg of H
2
​
SO
4
​
. Based on your knowledge of toichiometry, set up the table below to determine the amounts of each reactant and product after the reaction goes o completion..

Answer 1

To calculate the maximum grams of product in the reaction, construct a BCA table and determine the maximum grams of the possible product. The limiting reactant is Ca3(PO4)2, and the maximum moles of product formed is 6457.152 mol H3PO4. This corresponds to a maximum mass of 632,661.696 g.

Step-by-step explanation:

To determine the maximum grams of product in the reaction, we need to first construct a BCA (Before-Change-After) table. In this reaction, 1.00 kg of Ca3(PO4)2 and 1.00 kg of H2SO4 are reacted. Since the molar masses of Ca3(PO4)2 and H2SO4 are known, we can convert the masses to moles using their respective molar masses:

Ca3(PO4)2: 1.00 kg × (1000 g/kg) ÷ (310.177 g/mol) = 3228.576 mol

H2SO4: 1.00 kg × (1000 g/kg) ÷ (98.09 g/mol) = 10,184.103 mol

Next, we determine the stoichiometric ratios from the balanced chemical equation (3Ca3(PO4)2 + 10H2SO4 → 5CaSO4 + 6H3PO4).

Using the ratios, we find that the limiting reactant is Ca3(PO4)2, with a stoichiometric ratio of 3:1. Therefore, the maximum moles of product formed will be 3228.576 mol × (6 mol H3PO4 ÷ 3 mol Ca3(PO4)2) = 6457.152 mol H3PO4.

Finally, we convert the moles of H3PO4 to grams using the molar mass of H3PO4 (98.0 g/mol):

6457.152 mol × 98.0 g/mol = 632,661.696 g.