magnesium oxide reacts with phosphoric, H3PO4 TO PRODUCE MAGNESIUM PHOSPHATE AND WATER. HOW MANY GRAMS OF MAGNESIUM OXIDE ARE REQUIRED TO REACT COMPLETELY WITH 33.5G OF PHOSPORIC ACID?

To determine the amount of magnesium oxide required to react completely with 33.5 grams of phosphoric acid (H3PO4), we need to balance the chemical equation and use stoichiometry.

First, let's write the balanced chemical equation for the reaction between magnesium oxide (MgO) and phosphoric acid (H3PO4):

2MgO + H3PO4 -> Mg3(PO4)2 + H2O

According to the balanced equation, 2 moles of MgO react with 1 mole of H3PO4 to produce 1 mole of Mg3(PO4)2 and 1 mole of H2O.

Now, let's calculate the number of moles of H3PO4 based on the given mass of 33.5 grams. We need to convert grams to moles using the molar mass of H3PO4:

Molar mass of H3PO4 = 3(1.01 g/mol for H) + 1(15.99 g/mol for P) + 4(16.00 g/mol for O) = 98.00 g/mol

Number of moles of H3PO4 = 33.5 g / 98.00 g/mol ≈ 0.342 moles

According to the balanced equation, 2 moles of MgO react with 1 mole of H3PO4. Therefore, the number of moles of MgO required is twice the moles of H3PO4.

Number of moles of MgO = 2 * 0.342 moles = 0.684 moles

Finally, let's calculate the mass of MgO required by multiplying the number of moles by the molar mass of MgO:

Molar mass of MgO = 24.31 g/mol for Mg + 16.00 g/mol for O = 40.31 g/mol

Mass of MgO = 0.684 moles * 40.31 g/mol ≈ 27.56 grams

Therefore, approximately 27.56 grams of magnesium oxide are required to react completely with 33.5 grams of phosphoric acid.