Question

How many grams of sulfur are needed to react completely with 246 g of mercury to form hgs?

Answer 1

The reaction between mercury (Hg) and sulfur (S) to form HgS is:

Hg + S ------------- HgS

Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS

The given mass of Hg = 246 g

Atomic mass of Hg = 200.59 g/mol

# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles

Based on the reaction stoichiometry,

# moles of S that would react = 1.226 moles

Atomic mass of S = 32 g/mol

Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g

39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS

Answer 2

39.3 grams of sulfur reacted completely with 246 grams of mercury to form HgS.

FURTHER EXPLANATION

To solve this problem, first write the balanced chemical equation for the given reaction:

Hg + S → HgS

To solve the problem, we use the stoichiometric ratios of mercury and sulfur.

First, convert the mass of mercury to moles using the molar mass (200.59 g/mol).

$moles \ Hg = 246 \ g \ Hg \times \frac{1 \ mol \ Hg}{200.59 \ g}\\\\moles \ Hg = 1.226 \ mol$

Now, determine the moles of sulfur formed using the stoichiometric ratio of sulfur and mercury which is 1:1

$moles \ S \ = 1.226 \ mol \ Hg \times \frac{`1 \ mol \ S}{1 \ mol \ Hg} \\\\moles \ S = 1.226 \ mol$

To get the mass of sulfur that reacted with mercury to form HgS, use the molar mass of sulfur (32.06 g/mol)

$mass \ of \ S = 1.226 \ mol \ S \times \frac{32.06 \ g}{1 \ mol \ S}\\\\\boxed {mass \ of \ S = 39.31 \ g}$

Since the given only has 3 significant figures, the final answer must also have the same. Therefore,

$\boxed {\boxed {mass \ of \ S = 39.3 \ g}}$

Learn More

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Keywords: stoichiometry, dimensional analysis