Explanation:
To find the number of moles of phosphorus (P) in Ni₃(PO₄)₂, we need to consider the stoichiometry of the compound.
From the chemical formula, we can see that there are 2 moles of phosphorus (P) for every mole of Ni₃(PO₄)₂. So, if there are 1.34 moles of Ni, then the number of moles of P will be:
\[ \text{Number of moles of P} = 2 \times \text{Number of moles of Ni} = 2 \times 1.34 = 2.68 \text{ moles of P} \]
Now, to find the grams of phosphorus, we need to use the molar mass of phosphorus, which is approximately 30.97 g/mol.
\[ \text{Mass of P} = \text{Number of moles of P} \times \text{Molar mass of P} = 2.68 \times 30.97 = 82.97 \text{ grams of P} \]
Next, to find the total mass of the Ni₃(PO₄)₂ sample, we need to consider the molar masses of each element in the compound and sum them up.
\[ \text{Molar mass of Ni} = 58.69 \text{ g/mol} \]
\[ \text{Molar mass of P} = 30.97 \text{ g/mol} \]
\[ \text{Molar mass of O} = 15.999 \text{ g/mol} \]
\[ \text{Total molar mass of Ni₃(PO₄)₂} = (3 \times \text{Molar mass of Ni}) + (2 \times \text{Molar mass of P}) + (8 \times \text{Molar mass of O}) \]
\[ = (3 \times 58.69) + (2 \times 30.97) + (8 \times 15.999) \]
\[ = 176.07 + 61.94 + 127.992 \]
\[ = 365.002 \text{ g/mol} \]
Now, to find the total mass of the Ni₃(PO₄)₂ sample:
\[ \text{Total mass} = \text{Number of moles} \times \text{Molar mass} = 1.34 \times 365.002 = 488.753 \text{ grams} \]
So, the sample of Ni₃(PO₄)₂ containing 1.34 moles of Ni contains 2.68 moles of P, and its total mass is approximately 488.753 grams.
