Final answer:
To determine the mass of strontium phosphate formed from 6.25 grams of strontium chloride reacting with excess phosphoric acid, we first write the balanced chemical equation, convert the mass of SrCl2 to moles, use stoichiometry to find the moles of Sr3(PO4)2 produced, and then calculate its mass resulting in 5.93 grams of strontium phosphate.
Step-by-step explanation:
To calculate the mass of strontium phosphate that forms when 6.25 grams of strontium chloride react with excess phosphoric acid, we first need to write the balanced chemical equation for the reaction:
3 SrCl2 + 2 H3PO4 → Sr3(PO4)2 + 6 HCl
Next, we find the molar mass of SrCl2 (Sr = 87.62 g/mol, Cl = 35.45 g/mol) which is 87.62 g/mol + 2(35.45 g/mol) = 158.52 g/mol. Then, we convert the mass of SrCl2 to moles:
6.25 g SrCl2 x (1 mol SrCl2 / 158.52 g SrCl2) = 0.0394 mol SrCl2
According to the stoichiometry of the balanced equation, it takes 3 moles of SrCl2 to produce 1 mole of Sr3(PO4)2, thus:
0.0394 mol SrCl2 x (1 mol Sr3(PO4)2 / 3 mol SrCl2) = 0.0131 mol Sr3(PO4)2
Finally, we calculate the mass of Sr3(PO4)2 (Sr = 87.62 g/mol, P = 30.97 g/mol, O = 16.00 g/mol) which is 3(87.62 g/mol) + 2 (30.97 g/mol) + 8(16.00 g/mol) = 452.80 g/mol. So, the mass of Sr3(PO4)2 produced is:
0.0131 mol Sr3(PO4)2 x 452.80 g/mol = 5.93 g Sr3(PO4)2
Therefore, the mass of strontium phosphate formed is 5.93 grams.