How many grams of mercury are needed to react completely with 157g of sulfur to form hgs?

2 answers:

8 0

To solve this we need to balance the equations first.
So Hg + S --> HgS is balanced
One mole of Hg requires one mole of S to form one mole of HgS.
Number of moles of Sulphur = mass/ molar mass = 157/32 = 4.906
So 4.90 moles of S reacts with 4.90 moles of Hg.
Hence there are 4.90 moles of 4.90 of Hg.
Mass = number of moles * molar mass of Hg
Mass = 4.906 * 200.59 = 982.891g

joja [24]2 years ago

8 0

Answer:

983 g

Explanation:

Let's consider the following reaction.

Hg + S → HgS

The molar mass of S is 32.07 g/mol. The moles corresponding to 157 g are:

157 g × (1 mol/32.07 g) = 4.90 mol

The molar ratio of Hg to S is 1:1. The moles of Hg that react are 4.90 moles.

The molar mass of Hg is 200.59 g/mol. The mass corresponding to 4.90 moles is:

4.90 mol × (200.59 g/mol) = 983 g

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lianna [129]

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21.6 g

Explanation:

The reaction that takes place is:

CH₄ + 2O₂ → CO₂ + 2H₂O

First we convert the given masses of both reactants into moles, using their respective molar masses:

9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄

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Now we calculate how many moles of water are produced, using the number of moles of the limiting reactant: