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Answer:
Explanation:
Hello,
In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:
Thus, as only 8 g were actually yielded, we compute the percent yield:
Best regards.
Answer:
1. C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. V = 596L
Explanation:
Butane (C₄H₁₀) reacts with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O) thus:
C₄H₁₀ + O₂ → CO₂ + H₂O
1. The balanced chemical equation is:
C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O
2. 0,360kg of butane are:
360g×=<em>6,19moles of butane</em>
These moles of butane are:
6,19moles of butane×= <em>24,8 moles CO₂</em>
Using V=nRT/P
Where:
n are moles (24,8 moles CO₂); R is gas constant (0,082atmL/molK); T is temperature, 20°C (293,15K); and P is pressure (1atm).
Volume (V) is:
<em>V = 596L</em>
I hope it helps!