Mg + P4 → Mg3P2 Assuming an unlimited amount of phosphorous (P4), how many grams of Mg are needed in order to
produce 1200.00 grams of Mg3P2?

648.7g of Mg must be added

Explanation:

Based on the reaction:

3Mg + 1/2P4 → Mg3P2

3 moles of Mg produce 1 mole of Mg3P2

To solve this problem we neeed to find the moles of Mg3P2 in 1200g. With these moles and the chemical reaction we can finf moles of Mg and its mass:

Moles Mg3P2 -Molar mass: 134.88g/mol-:

1200.0g Mg3P2 * (1mol / 134.88g) = 8.897 moles of Mg3P2

Moles Mg:

8.897 moles of Mg3P2 * (3mol Mg / 1mol Mg3P2) = 26.69 moles of Mg

Mass Mg -Molar mass: 24.305g/mol-:

26.69 moles of Mg * (24.305g / 1mol) =