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2 years ago

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

Chemistry

1 answer:

Maksim231197 [3]2 years ago

7 0

The theoretical yield of $SnS_2$ will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of $SnBr_4$ to $Na_2S$ is 1:2.

Mole of 48.1 mL, 0.478 M $SnBr_4$ = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M $Na_2S$ = 0.160 x 48.8/1000 = 0.0078 moles

$SnBr_4$ $Na_2S$ is the limiting reactant.

Mole ratio of $SnBr_4$ and $SnS_2$ = 1:1

Equivalent mole of $SnS_2$ = 0.023 moles

Mass of 0.023 noles $SnS_2$ = 0.023 x 182.81 = 4.20 grams

With 0.0333 g of $SnS_2$ recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

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Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

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The best equipment would be the graduated cylinder. Why?

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4 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.10 g of ethane is

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Answer:

Ethane is the limiting reactant, so no mass of ethane will be left over by the chemical reaction. All the mass will react, the 2.1 grams of ethane.

Explanation:

First of all, we need to determine the reaction and the limiting reactant to work with the stoichiometry.

The equation is: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

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2.10 g / 30 g/mol = 0.07 moles of ethane

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To determine the limiting reactant, we start with oxygen:

7 moles of O₂ can react with 2 moles of ethane

Then, 0.375 moles of O₂ will react with (0.375 . 2) / 7= 1.31 moles of ethane.

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3 years ago

Read 2 more answers

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

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