Question

How many grams of ba3(PO₄)2 can be formed when 126g of barium hydroxide (171.34g/mol) reacts with excess h3PO₄?

1) 1327.9g ba3(PO₄)2
2) 11.9g ba3(PO₄)2
3) 442.6g ba3(PO₄)2
4) 147.5g ba3(PO₄)2
5) 221.3g ba3(PO₄)2

Answer 1

To find the grams of Ba3(PO4)2 formed, we calculate the moles of Ba(OH)2, use the mole ratio between Ba(OH)2 and Ba3(PO4)2, and convert moles to grams using the molar mass of Ba3(PO4)2. The correct answer is 4) 147.5g Ba3(PO4)2.

Step-by-step explanation:

To determine the number of grams of Ba3(PO4)2 that can be formed, we first need to calculate the moles of Ba(OH)2 using the given mass and molar mass:

Moles of Ba(OH)2 = Mass of Ba(OH)2 / Molar mass of Ba(OH)2

Moles of Ba(OH)2 = 126g / 171.34g/mol = 0.7356 mol

According to the balanced equation, the ratio between Ba(OH)2 and Ba3(PO4)2 is 3:1. So, the number of moles of Ba3(PO4)2 formed will be:

Moles of Ba3(PO4)2 = 0.7356 mol * (1 mol Ba3(PO4)2 / 3 mol Ba(OH)2) = 0.2452 mol

To convert moles to grams, we use the molar mass of Ba3(PO4)2:

Mass of Ba3(PO4)2 = Moles of Ba3(PO4)2 * Molar mass of Ba3(PO4)2

Mass of Ba3(PO4)2 = 0.2452 mol * 601.93g/mol = 147.5g

Therefore, the correct answer is option 4) 147.5g Ba3(PO4)2.