Given 32.6 g of Ba(OH)2 and 45.4 g of H3PO4, how many grams of Ba3(PO4)2 can you

Question

Given 32.6 g of Ba(OH)2 and 45.4 g of H3PO4, how many grams of Ba3(PO4)2 can you produce, assuming the reaction goes to completion? Please show steps.

Explanation / Answer

First, write a balanced chemical equation: 3Ba(OH)2 + 2H3PO4 --> Ba3(PO4)2 + 6H2O First, solve for the moles of Ba(OH)2 and H3PO4: 32.6 g x 1 mol/171.34 g = 0.190 moles of Ba(OH)2 45.4 g x 1 mol/98.00 g = 0.463 moles of H3PO4 Now, solve for the moles of Ba3(PO4)2 from the moles of Ba(OH)2 and H3PO4 using the balanced chemical equation: 0.190 moles of Ba(OH)2 x 1 mole of Ba3(PO4)2/3 moles of Ba(OH)2 = 0.0633 moles of Ba3(PO4)2 0.463 moles of H3PO4 x 1 mole of Ba3(PO4)2/2 moles of H3PO4 = 0.232 moles of Ba3(PO4)2 Since Ba(OH)2 produces the fewer moles of Ba3(PO4)2, it is the limiting reagent. Mass of Ba3(PO4)2: 0.0633 moles x 601.93 g/mol = 38.1 g Hope this helps! :)

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.