Question

How many grams in 6.58 x 10^25 formula units of nickel (II) phosphate

Answer 1

`\boxed{\text{40 000 g}}`

Step-by-step explanation:

We must convert formula units of Ni₃(PO₄)₂ to moles and then to grams.

Step 1. Convert formula units to moles

`\text{Moles of Ni$_(3)$(PO$_(4)$)$_(2)$}\\= 6.58 *10^(25)\text{ formula units Ni$_(3)$(PO$_(4)$)$_(2)$} * \frac{\text{1 mol Ni$_(3)$(PO$_(4)$)$_(2)$}}{6.022 *\ 10^(23) \text{ formula units Ni$_(3)$(PO$_(4)$)$_(2)$}}\\\\= \text{109.3 mol Ni$_(3)$(PO$_(4)$)$_(2)$}`

Step 2. Convert moles to grams

`\text{Mass of Ni$_(3)$(PO$_(4)$)$_(2)$}\\\\= \text{109.3 mol Ni$_(3)$(PO$_(4)$)$_(2)$} * \frac{\text{366.02 g Ni$_(3)$(PO$_(4)$)$_(2)$}}{\text{1 mol Ni$_(3)$(PO$_(4)$)$_(2)$}}\\\\= \text{378 g Ni$_(3)$(PO$_(4)$)$_(2)$}\\\text{The mass of Ni$_(3)$(PO$_(4)$)$_(2)$ is } \boxed{\textbf{40 000 g}}`