Suppose you need to prepare 138.1 mL of a 0.190 M aqueous solution of NaCl. What mass, in grams, of NaCl do you need to use to m
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Answer:
30.0 mol CO₂
Explanation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
To answer this problem we need to convert moles of C₃H₈ into moles of CO₂: We'll do that by using the <u>stoichiometric coefficients</u>, using a conversion factor that has C₃H₈ moles in the denominator and CO₂ moles in the numerator:
10.0 mol C₃H₈ * = 30.0 mol CO₂