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Ludmilka [50]
3 years ago
15

Part B When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 22.8 g of carbon were burned in the pres

ence of 73.8 g of oxygen, 13.0 g of oxygen remained unreacted. What mass of carbon dioxide was produced? Express your answer to one decimal place and include the appropriate units. View Available Hint(s)
Chemistry
1 answer:
aksik [14]3 years ago
6 0

<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

<u>Explanation:</u>

As, some amount of oxygen gas is left after the reaction is completed. So, it is present in excess and is considered as an excess reagent.

Thus, carbon is considered as a limiting reagent because it limits the formation of product.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of carbon = 22.8 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon}=\frac{22.8g}{12g/mol}=1.9mol

The chemical equation for the reaction of carbon and oxygen gas follows:

C+O_2\rightarrow CO_2

By Stoichiometry of the reaction:

1 mole of carbon produces 1 mole of carbon dioxide gas

So, 1.9 moles of carbon will produce = \frac{1}{1}\times 1.9=1.9moles of carbon dioxide gas

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.9 moles

Putting values in equation 1, we get:

1.9mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.9mol\times 44g/mol)=83.6g

Hence, the amount of carbon dioxide gas produced in the reaction is 83.6 grams

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12. What is the volume of 0.07 mol of neon gas at STP?
scoundrel [369]
<h3>Answer:</h3>

2 L Ne

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.07 mol Ne (g)

<u>Step 2: Identify Conversions</u>

STP - 22.4 L per mole

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 0.07 \ mol \ Ne(\frac{22.4 \ L \ Ne}{1 \ mol \ Ne})
  2. Multiply:                              \displaystyle 1.568 \ L \ Ne

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1.568 L Ne ≈ 2 L Ne

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Which expression is a measurement of density? 20 g/cm 2 5.2 cm 3 0.71 mg 5.13 kg/L
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Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8
kaheart [24]

Answer :  The average rate of consumption of H^+ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

As we are given:

\frac{d[Br^-]}{dt}=1.81M/s

Now we have to determine the average rate of consumption of H^+ during the same time interval.

As,

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

or,

-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times (1.81M/s)

\frac{d[H^+]}{dt}=2.17M/s

Thus, the average rate of consumption of H^+ during the same time interval is, 2.17 M/s

6 0
4 years ago
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