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zheka24 [161]
2 years ago
13

How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

Chemistry
1 answer:
zepelin [54]2 years ago
4 0
<h3>Answer:</h3>

0.34 mol S

<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>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

11 g S

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

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

  1. Set up:                                \displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})
  2. Multiply/Divide:                  \displaystyle 0.343 \ mol \ S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

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Answer:

13.7 moles of O₂ are needed

Explanation:

In order to find the moles of reactants that may react to make the products we need to determine the reaction:

Reactants are hydrogen and oxygen

Product: Water

2 moles of hydrogen can react to 1 mol of oxygen and produce 2 moles of water.

Balanced reaction: 2H₂(g) + O₂(g) →  2H₂O(l)

If 2 moles of hydrogen need 1 mol of oxygen to react

Therefore, 27.4 moles of H₂ must need (27.4 .1) / 2 = 13.7 moles of O₂

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3 years ago
N2 + 3H2 mc011-1.jpg 2NH3<br> What is the mole ratio of hydrogen to ammonia?
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1 N₂ + 3 H₂ = 2 NH₃

Mole  ratio <span> of hydrogen to ammonia :

3 moles H</span>₂  :  2 moles NH₃   or  3 : 2

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3 years ago
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Methanol can be synthesized by combining carbon monoxide and hydrogen. CO(g) 2H2(g) CH3OH(g) . A reaction vessel contains the th
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Answer:

The answer is C. The partial pressure of hydrogen will be unchanged.

Explanation:

CO_{(g)} + 2H_{2(g)}  ⇒  CH_{3}OH

Argon with electronic configuration 1s^{2}  2s^{2}  2p^{6}  3s^{2}  3p^{6} (that is atomic number 18) is an inert gas making it unreactive and it's addition to the reaction has no effect on the partial pressure of either the reactant or production or the state of the system.

The partial pressure of hydrogen will remain unchanged on the addition of Argon.

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The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia: 2N0(g) +02(g) 2NO2 (g) The
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The question is incomplete, complete question is :

The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols

Answer:

Equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

Explanation:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}

Step 2 :  2NO(g) +O_2(g) \rightleftharpoons 2NO_2 (g)

Expression of an equilibrium constant can be written as:

K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}

Multiply and divide [NO]^4;

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}

K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}

K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}

K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2

K=K_1\time (K_2)^2

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

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