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How many moles of PbS are equivalent to 458 g of PbS?

Molar mass of PbS = 239.27 g/mol

[?] mol PbS

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To determine how many moles of PbS (Lead(II) sulfide) are equivalent to a given mass of 458 grams, we must use the following relationship between mass, molar mass, and moles:

[tex]\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \][/tex]

Given data:
- Mass of PbS: 458 grams
- Molar mass of PbS: 239.27 grams per mole

Here's the step-by-step process:

1. Write down the formula:
[tex]\[ \text{Moles of PbS} = \frac{\text{Mass of PbS}}{\text{Molar Mass of PbS}} \][/tex]

2. Substitute the given values into the formula:
[tex]\[ \text{Moles of PbS} = \frac{458 \text{ grams}}{239.27 \text{ grams per mole}} \][/tex]

3. Perform the division:
[tex]\[ \text{Moles of PbS} = 1.9141555564843062 \text{ moles} \][/tex]

Therefore, the number of moles of PbS equivalent to 458 grams is approximately [tex]\( 1.9142 \)[/tex] moles, when rounded to four decimal places.
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