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------------------------------------------------ How many grams of phosphorus (P₄) react with 35.5 L of O₂ at STP to form solid tetraphosphorus decaoxide?

To find the mass of phosphorus that reacts with 35.5L of O2 to form tetraphosphorus decaoxide at STP, we calculate the moles of O2, use the molar ratio to find moles of P4 needed, and convert to grams, resulting in 157.9 grams of P4.

Explanation:

To determine how many grams of phosphorus (P4) react with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide (P4O10), we first need to use the chemical reaction stoichiometry. The balanced chemical equation is:

4 P4 + 5 O2 ightarrow 2 P4O10

At STP (Standard Temperature and Pressure), 1 mole of a gas occupies 22.4 liters. Therefore, 35.5 liters of O2 is:

35.5 L O2 / 22.4 L/mol = 1.5857 moles of O2

Using the molar ratio from the balanced equation, we can determine the moles of P4 needed:

(1.5857 moles of O2) * (4 moles of P4 / 5 moles of O2) = 1.2686 moles of P4

The molar mass of P4 is 4 times the atomic mass of phosphorus (approximately 30.974 g/mol), so:

1.2686 moles of P4 * (4 * 30.974 g/mol) = 157.9 grams of P4

Therefore, 157.9 grams of phosphorus (P4) reacts with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide.