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sleet_krkn [62]
3 years ago
5

The percent yield of this reaction is consistently 92.0%. CH4(g) + 4 S(g) → CS2(g) + 2 H2S(g) How many grams of sulfur would be

needed to obtain 72.57 g of CS2?
Chemistry
1 answer:
dybincka [34]3 years ago
6 0

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

    CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S  ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of  CS₂ .

since ,

Moles of  CS₂  = given mass of CS₂ / Molecular weight  of  CS₂

Since ,

the Molecular weight of  CS₂ = 76

Given ,  mass of CS₂ =  72.57 g

Moles of  CS₂ = 72.57  / 76 = 0.95 mol

Since ,

The yield is 92.0 % .

Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles

Mass of S required = 4.13 * 32 = 132.17 g .

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AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

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If you have 20.0 g of CO2, how many atoms of Oxygen (O) are present in the sample?
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Answer:

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Explanation:

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