Answer :
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!
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!