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Ghella [55]
3 years ago
9

How many O2 are needed to react with 8.95 moles of Al?​

Chemistry
2 answers:
Svetlanka [38]3 years ago
7 0

Answer:

The answer to your question is 6.713 moles of O₂ or 107.4 grams

Explanation:

Data

moles and grams of O₂

moles of Al = 8.95

Balanced chemical reaction

                4Al  +  3 O₂   ⇒    2Al₂O₃

Process

1.- Use proportions and cross multiplication to find the answer. Use the coefficients of the balanced reaction.

               4 moles of Al ---------------- 3 moles of O₂

                8.95 moles  ----------------- x

                x = (8.95 x 3) / 4

                x = 26.85 / 4

               x = 6.713 moles of O₂

2.- Calculate the grams of O₂

                 16 g of O₂ ------------------ 1 mol

                   x               ----------------  6.713 moles

                  x = (6.713 x 16) / 1

                 x = 107.4 grams

Over [174]3 years ago
7 0

Answer:

We need 6.71 moles of O2 (or 214.7 grams of O2) to react with 8.95 moles of Al

Explanation:

Step 1: Data given

Number of moles of Al = 8.95 moles

Step 2: The balanced equation

4Al + 3O2 → 2Al2O3

Step 3: Calculate moles number of moles O2 needed to react with Al

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 8.95 moles Al we need 3/4 * 8.95 = 6.71 moles O2

There will 8.95/2 = 4.48 moles Al2O3 be produced

We need 6.71 moles of O2 to react with 8.95 moles of Al

Mass O2 = moles O2 * molar mass O2

Mass O2 = 6.71 moles * 32.0 g/mol

Mass O2 =214.7 grams O2

We need 214.7 grams of O2 to react with 8.95 moles Al

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8 0
3 years ago
What is the final volume of a gas (in liters) when the initial volume is 14.00 L at
8_murik_8 [283]

Answer:

V_2=1344L

Explanation:

Hello there!

In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:

P_2V_2=P_1V_1\\\\V_2=\frac{P_2V_2}{P_1}\\

Then, we plug in the initial and final pressures and the initial volume to obtain:

V_2=\frac{14.00L*0.9600atm}{0.01000atm}\\\\V_2=1344L

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4 0
2 years ago
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3 years ago
URGENT!!-- Please help!
blondinia [14]

Moles of gas = 0.369

<h3>Further explanation</h3>

Given

P = 2 atm

V = 5.3 L

T = 350 L

Required

moles of gas

Solution

Ideal gas Law

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{2\times 5.3}{0.082\times 350}\\\\n=0.369

Avogadro's law : at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles  

moles of O₂ = 45% x 0.369 = 0.166

moles of Ar = 12% x 0.369 = 0.044

moles of N = 43% x 0.369 = 0.159

3 0
2 years ago
Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen
Oksanka [162]
<h3>Answer:</h3>

10.6 mol NO

<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> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 4NH₃ + 5O₂ → 4NO + 6H₂O

[Given] 13.2 mol O₂

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

[RxN] 5 mol O₂ → 4 mol NO

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 13.2 \ mol \ O_2(\frac{4 \ mol \ NO}{5 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 10.56 \ mol \ NO

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

10.56 mol NO ≈ 10.6 mol NO

5 0
2 years ago
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