Question

How many moles of phosphorous are in 0.25 g of Na3PO4?

Answer 1

0.001525 mol

Step-by-step explanation:

To find the number of moles of phosphorous in 0.25 g of Na₃PO₄, we will first calculate the molar mass of Na₃PO₄, then use stoichiometric ratios based on its molecular formula, and finally apply the given mass to determine the moles of phosphorous.

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Step (1): Determine the molar mass of Na₃PO₄

1 mole of Na₃PO₄ consists of:

• 3 moles of Na (Sodium)
• 1 mole of P (Phosphorus)
• 4 moles of O (Oxygen)

Atomic weights:

• Na (Sodium) = 22.990 g/mol
• P (Phosphorus) = 30.974 g/mol
• O (Oxygen) = 15.999 g/mol

Using atomic weights to determine the molar mass of Na₃PO₄:

`\Longrightarrow \text{Molar Mass of $Na_2SO_4$}=(3*22.990)(1*30.974)(4*15.999)\\\\\\\\\therefore \text{Molar Mass of $Na_2SO_4$}=163.94 \ g/mol`

Step (2): Using stoichiometry

From the molecular formula, Na₃PO₄, it is evident that:

1 mole of Na₃PO₄ contains 1 mole of Phosphorus (P)

Step (3): Calculate the moles of phosphorus from 0.25 g of Na₃PO₄

Using the formula:

`\text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}}`

`\Longrightarrow \text{Number of moles of $Na_2SO_4$}=(0.25 \ g)/(163.94 \ g/mol)\\\\\\\\\therefore \boxed{\text{Number of moles of $Na_2SO_4$}=0.001525 \ mol}`

There are 0.001525 moles of phosphorous in 0.25 g of Na₃PO₄.