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Lerok [7]
4 years ago
11

In Part A, you found the amount of product (2.60 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

t B, you found the amount of product (2.20 mol P2O5 ) formed from the given amount of oxygen and excess phosphorus.Now, determine how many moles of P2O5 are produced from the given amounts of phosphorus and oxygen.Express your answer to three significant figures and include the appropriate units.
Chemistry
1 answer:
m_a_m_a [10]4 years ago
3 0

Explanation:

Equation of the reaction:

P₄ + 5O₂ → 2P₂O₅.

Given:

Mass of P₄ = 161 g.

Molar mass = 124 g/mol

Number of moles of P₄ = mass/molar mass

n(P₄) = 161 g ÷ 124 g/mol.

= 1.3 mol.

Mass of O₂ = 176 g.

Molar mass of O₂ = 2 ×16

= 32 g/mol

Number of moles = 176 g ÷ 32 g/mol.

= 5.5 moles

Calculating the limiting reactant,

5.5 mol of O2/5 mole of O2 × 1 mole of P₄

= 2.75 moles of P₄ (> 1.3 moles)

P₄ is the limiting reagent.

By stoichiometry, 1 mole of P₄ gives 2 mole of P₂O₅

Therefore, number of moles of P₂O₅ = 2 × 1.3

= 2.6 moles of P₂O₅ was formed.

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<em>A biochemist carefully measures the molarity of magnesium ion in 47, mL of cell growth medium to be 97 uM. Unfortunately, a careless graduate student forgets to cover the container of growth medium and a substantial amount of the solvent evaporates. The volume of the cell growth medium falls to 6.0 mL. Calculate the new molarity of magnesium ion in the cell growth medium Be sure your answer has the correct number of significant digits.</em>

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