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Question
fill in the blank with the letter next to the path you took.
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a. g c₃h₆ → l c₃h₆ → l h₂
b. g c₃h₆ → mol c₃h₆ → mol h₂ → g h₂
c. g c₃h₆ → g h₂ → l h₂
d. g c₃h₆ → mol c₃h₆ → l c₃h₆ → l h₂
e. g c₃h₆ → mol c₃h₆ → mol h₂ → l h₂
To determine the correct path, we analyze the stoichiometric process. We start with grams of \( \text{C}_3\text{H}_6 \), convert to moles of \( \text{C}_3\text{H}_6 \) (using molar mass), then use the mole ratio to find moles of \( \text{H}_2 \), and finally convert moles of \( \text{H}_2 \) to liters (if dealing with gas volume, assuming STP or relevant conditions for volume - mole conversion). Option E follows this: \( \text{g}\ \text{C}_3\text{H}_6 \to \text{mol}\ \text{C}_3\text{H}_6 \to \text{mol}\ \text{H}_2 \to \text{L}\ \text{H}_2 \). Option A skips mole - based stoichiometry. Option B converts to grams of \( \text{H}_2 \) instead of liters. Option C has an incorrect direct conversion from \( \text{g}\ \text{C}_3\text{H}_6 \) to \( \text{g}\ \text{H}_2 \) without mole ratio. Option D includes an unnecessary conversion to liters of \( \text{C}_3\text{H}_6 \).
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E. \( \text{g}\ \text{C}_3\text{H}_6 \to \text{mol}\ \text{C}_3\text{H}_6 \to \text{mol}\ \text{H}_2 \to \text{L}\ \text{H}_2 \)