Phosphoric acid is very important compound used to make fertilizers.PPhosphoric acid can be prepared in two ways: Reaction 1: P4 + 5O2 = P4O10. Reaction 2: P4O10 + 6H29 = 4H3PO4. Supposing we allow 272 g of phosphorus to react with excess oxygen which forms tetra phosphorus decoxide in 96.8% yield. In the second step reaction, a 96.8% yield of H3PO4 is obtained. Work out the mass of H3PO4 obtained?

Produced mass of H3PO4 =62/4×31×0.85×0.9×4×98=149.94g

To calculate the mass of H3PO4 obtained, we need to use stoichiometry and the given information to determine the moles of phosphorus and phosphoric acid. Then, we can convert the moles of phosphoric acid to grams.

1. Calculate the moles of P4O10:
The balanced equation for Reaction 1 shows that 1 mole of P4 reacts with 5 moles of O2 to produce 1 mole of P4O10.
From the balanced equation, the molar ratio between P4 and P4O10 is 1:1.
So, the moles of P4O10 formed can be calculated as:
moles of P4O10 = mass of P4O10 / molar mass of P4O10

Given:
mass of P4O10 = 272 g
molar mass of P4O10 = 4 * (30.97 g/mol) + 10 * (16.00 g/mol) = 283.88 g/mol

moles of P4O10 = 272 g / 283.88 g/mol = 0.958 mol

2. Calculate the moles of H3PO4:
The balanced equation for Reaction 2 shows that 1 mole of P4O10 reacts with 6 moles of H2O to produce 4 moles of H3PO4.
From the balanced equation, the molar ratio between P4O10 and H3PO4 is 1:4.
So, the moles of H3PO4 formed can be calculated as:
moles of H3PO4 = 4 * moles of P4O10 (using the 96.8% yield)

moles of H3PO4 = 4 * 0.958 mol = 3.832 mol

3. Calculate the mass of H3PO4:
To determine the mass of H3PO4 obtained, we need to know the molar mass of H3PO4.

molar mass of H3PO4 = 3 * (1.01 g/mol) + 1 * (15.999 g/mol) + 4 * (1.008 g/mol) = 98.00 g/mol

mass of H3PO4 = moles of H3PO4 * molar mass of H3PO4
mass of H3PO4 = 3.832 mol * 98.00 g/mol = 375.94 g

Therefore, the mass of H3PO4 obtained is approximately 375.94 grams.