To determine the mass of phosphorus used up when 60.0 grams of sodium reacts completely in the provided chemical reaction, let's go through the problem step by step:
1. Chemical Equation:
[tex]\[
3 \text{ Na} + \text{ P} \xrightarrow{\Delta} \text{ Na}_3\text{P}
\][/tex]
According to this balanced chemical equation, 3 moles of sodium (Na) react with 1 mole of phosphorus (P) to produce sodium phosphide (Na[tex]\(_3\)[/tex]P).
2. Convert Given Mass to Moles:
The molar mass of Na is given as 22.99 g/mol.
We have 60.0 grams of Na. To find the moles of Na:
[tex]\[
\text{Moles of Na} = \frac{\text{mass of Na}}{\text{molar mass of Na}} = \frac{60.0 \text{ grams}}{22.99 \text{ g/mol}} \approx 2.6098303610265337 \text{ moles}
\][/tex]
3. Stoichiometric Conversion (Moles of Na to Moles of P):
According to the balanced equation, 3 moles of Na react with 1 mole of P. Therefore, we can find the moles of P required:
[tex]\[
\text{Moles of P} = \frac{\text{Moles of Na}}{3} = \frac{2.6098303610265337 \text{ moles}}{3} \approx 0.8699434536755112 \text{ moles}
\][/tex]
4. Convert Moles of P to Mass of P:
The molar mass of P is given as 30.97 g/mol.
Now, we can find the mass of P:
[tex]\[
\text{Mass of P used} = \text{moles of P} \times \text{molar mass of P} = 0.8699434536755112 \text{ moles} \times 30.97 \text{ g/mol} \approx 26.942148760330582 \text{ grams}
\][/tex]
So, the mass of phosphorus used up in the reaction is approximately 26.94 grams.
If you have any further questions or need any additional clarifications, feel free to ask!
