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3 years ago

2. Identify the limiting reactant when 4.68 g of iron reacts with 2.88 g of sulfur to produce Fes.

Chemistry

1 answer:

Vinvika [58]3 years ago

6 0

Answer:

Iron is limiting reactant

Explanation:

Based on the reaction:

Fe + S → FeS

1 mole of iron reacts per mole of Sulfur

To solve this question we must convert the mass of each reactant to moles using molar masses of each reactant. As the reaction is 1:1, the reactant with the lower amount of moles is limiting reactant.

Moles Fe -Molar mass: 55.845g/mol-

4.68g * (1mol / 55.845g) = 0.0838 moles

Moles S -Molar mass: 32.065g/mol-

2.88g * (1mol / 32.065g) = 0.0898 moles

As the amount of moles of Fe < Moles S,

Iron is limiting reactant

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Korolek [52]

Empirical formula is the simplest ratio of whole numbers of components in a compound
calculating for 100 g of compound
N O
mass 46.7 g 53.3 g
number of moles 46.7 g / 14 g/mol 53.3 g / 16 g/mol
= 3.33 = 3.33
divide by the least number of moles
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number of atoms
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O - 1
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Nesterboy [21]

Balanced chemical equation for the reaction is:

2S $O_{2}$ (g) + $O_{2}$ (g)+ 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

Moles of $H_{2} S_{} O_{4}$ formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of S $O_{2}$ = 11.5 moles

moles of $H_{2} S_{} O_{4}$ = ?

Moles of $O_{2}$ needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2S $O_{2}$ (g) + $O_{2}$ (g) + 2 $H_{2}$ O (l) ⇒ $H_{2} S_{} O_{4}$

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of S $O_{2}$ , x moles of $H_{2} S_{} O_{4}$ is formed

$\frac{1}{2} =\frac{x}{11.5}$

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

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