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lara [203]
3 years ago
12

2. Consider the following reaction: 4 FeS2 + 11O2 → 2 Fe2O3 + 8 SO2

Chemistry
1 answer:
belka [17]3 years ago
6 0
<h3>Answer:</h3>

Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

<h3>Explanation:</h3>

From the questions we are given;

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(s)

  • Moles of FeS₂ are 26.62 moles
  • Moles of Oxygen, O₂ are 59.44 moles

We are supposed to determine the limiting and excess reactants;

  • From the equation of the reaction given; 4 moles of FeS₂ required 11 moles of Oxygen gas.

Working with the amount of reactants given;

  • 26.62 moles of FeS₂ will require 73.205 moles of O₂ and only 59.44 moles of O₂ are available.
  • On the other hand 59.44 moles of O₂ requires 21.615 moles of  FeS₂, and we are given 26.62 moles of FeS₂ which means FeS₂ is in excess.

Conclusion;

We can conclude that Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

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- How many moles of H2O are required to react to form 2.5 grams of CH ?
abruzzese [7]

Answer:

0.312 moles of H2O

Explanation:

no. of moles of ch4= mass ÷ molar mass

                               =2.5 ÷ 16.04

                               =0.156 moles of ch4

According to balanced chemical equation

CH4        :        H2O

1 mole     :        2 moles

0.156 moles :       x moles  

by cross multiplication

x=  (0.156x2) ÷ 1

 = 0.312 moles of H2O

7 0
3 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

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

0.424086 J/g °C ≈ 0.424 J/g °C

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