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1 year ago

If 60g of rust (Fe2O₁) are made in the reaction below, how many Liters of oxygen must be used?

Chemistry

1 answer:

Aleksandr [31]1 year ago

7 0

Answer:

18.

Explanation:

1) according to the reaction (M(Fe)=56; M(O₂)=32; V₀=22.4):

4Fe+3O₂⇒2Fe₂O₃; - ν(O₂)=0.75*ν(Fe);

2) ν(Fe)=m(Fe)/M(Fe)=60/56≈1.071 [mol];

3) ν(O₂)=0.75*1.071=0.804 [mol];

4) V(O₂)=V₀*ν(O₂); ⇒ V(O₂)=22.4*0.804=18 [lit].

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CH4 + O2 → CO2 + H2O

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Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Organic

Naming carbons

Stoichiometry

Analyzing reaction rxn
Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

Step 2: Identify Conversions

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] Molar Mass of H: 1.01 g/mol

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

Step 3: Stoichiometry

[DA] Set up conversion: $\displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

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