Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14
The amount of Al required will be 15.77 grams
<h3>Stoichiometric problem</h3>
First, the equation of the reaction:

The mole ratio is 2:3.
Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles
Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles
Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
3 or more simple sugars combined
Explanation:
A polysaccharid molecules consist of a number of sugar molecules bonded together.
The amount remaining at the end of 5 half-lives is 7.81×10¹³ g
From the question given above, the following data were obtained:
- Half-life (t½) = 5730 years
- Original amount (N₀) = 2.5×10¹⁵ g
- Number of half-lives (n) = 5
- Amount remaining (N) =?
The amount remaining can be obtained as follow:
N = 1/2ⁿ × N₀
N = 1/2⁵ × 2.5×10¹⁵
N = 1/32 × 2.5×10¹⁵
N = 0.03125 × 2.5×10¹⁵
N = 7.81×10¹³ g
Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g
Learn more about half-life: brainly.com/question/25783920