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Vanyuwa [196]
3 years ago
8

Gaseous ethane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yiel

d of water formed from the reaction of of ethane and of oxygen gas? Be sure your answer has the correct number of significant digits in it.
Chemistry
1 answer:
Effectus [21]3 years ago
8 0

Answer:

1) Write the balanced equation:

2C2H6 + 7O2 ---> 4CO2 + 6H2O

2) Determine limiting reagent:

C2H6 ⇒ 13.8 g / 30.0694 g/mol = 0.45894 mol

O2 ⇒ 45.8 g / 31.9988 g/mol = 1.4313 mol

C2H6 ⇒ 0.45894 / 2 = 0.22947

O2 ⇒ 1.4313 / 7 = 0.20447

Oxygen is limiting.

3) Determine theoretical yield of water:

The oxygen : water molar ratio is 7 : 6

7 is to 6 as 1.4313 mol is to x

x = 1.2268286 mol of water

4) Convert moles of water to grams:

1.2268286 mol times 18.015 g/mol = 22.1 g (to three sig figs)

Solution to (b):

14.2 g / 22.1 g = 64.2%

Explanation:

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Explanation:

A reaction quotient is defined as the ratio of concentration of products over reactants raised to the power of their stoichiometric coefficients.

A reaction quotient is denoted by the symbol Q.

For example, 2Fe^{3+}(aq) + Zn(s) \rightleftharpoons 2Fe^{2+}(aq) + Zn^{2+}(aq)

The reaction quotient for this reaction is as follows.

            Q = \frac{[Fe^{2+}]^{2}[Zn^{2+}]}{[Fe^{3+}]^{2}}

[Zn] will be equal to 1 as it is present in solid state. Therefore, we don't need to write it in the reaction quotient expression.

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You're provided a 0.50 M solution of HCl. How many moles of HCl are present in 25 ml of this solution? ​
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0.0125 moles of HCl

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15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what
s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem.  </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:        52.00   80.91       291.71

                2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g:  15.00    15.00  

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

  Moles of Cr = 15.00 × 1/52.00

  Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×  

<em>Step 3</em>. Identify the<em> limiting reactant</em>  

Calculate the moles of CrCl₃ we can obtain from each reactant.  

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

   Theoretical yield of H₂ = 0.1854 × 3/6

   Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

       % Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

       % yield = 15.35/18.03 × 100

       % yield = <em>85.15 % </em>

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3 years ago
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Answer:

2,7,4,6

Explanation:

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