_H3PO4+_KOH->_K3PO4+_H2O

if 49.0 g of H3PO4 is reacted with excess KOH, determind the percent yield of K3PO4 if you produce 49.0 g of K3PO4.

See your Al2(SO4)3 problem.

To determine the percent yield of K3PO4, we need to compare the amount of K3PO4 obtained (49.0 g) with the theoretical yield of K3PO4, which is the maximum amount that could be obtained under ideal conditions.

First, we need to calculate the molar mass of H3PO4 and K3PO4.

1. H3PO4:
The molar mass of H is 1 g/mol, P is 31 g/mol, and O is 16 g/mol. Since there are three H atoms and four O atoms in H3PO4, the molar mass is:
(3 * 1 g/mol) + 31 g/mol + (4 * 16 g/mol) = 98 g/mol

2. K3PO4:
The molar mass of K is 39 g/mol, P is 31 g/mol, and O is 16 g/mol. Since there are three K atoms, one P atom, and four O atoms in K3PO4, the molar mass is:
(3 * 39 g/mol) + 31 g/mol + (4 * 16 g/mol) = 212 g/mol

Next, we can calculate the number of moles of H3PO4 reacted:

Moles of H3PO4 = Mass / Molar mass
Moles of H3PO4 = 49.0 g / 98 g/mol = 0.5 mol

According to the balanced chemical equation, the mole ratio between H3PO4 and K3PO4 is 1:1. Therefore, the number of moles of K3PO4 produced is also 0.5 mol.

We can now calculate the theoretical yield of K3PO4:

Theoretical yield = Moles of K3PO4 * Molar mass of K3PO4
Theoretical yield = 0.5 mol * 212 g/mol = 106 g

Now, we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (49.0 g / 106 g) * 100 = 46.2%

Therefore, the percent yield of K3PO4 is 46.2% when 49.0 g of H3PO4 is reacted with excess KOH and results in 49.0 g of K3PO4.