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Propionoic acid, C3H6O2, reacting with methanol, CH4O, will produce C4H8O2 and water. If 70.0 g of propionoic acid and 60.0 g of methanol are reacted together, which one will be the limiting reactant?

C3H6O2 + CH4O → C4H8O2 + H2O

Sagot :

thisisbenmanley
First, find the molar masses of each substance.

C3H6O2 = 74 g/mol

CH4O = 32 g/mol

Then, use the molar mass to find how many moles of each are present:

C3H6O2:  [tex](70.0g\ C_{3}H_{6}O_{2})(\frac{1 mol\ C_{3}H_{6}O_{2}}{74.0 g\ C_{3}H_{6}O_{2}}) = 0.945945945 mol\ C_{3}H_{6}O_{2}[/tex]

CH4O:  [tex](60.0g\ CH_{4}O)(\frac{1mol\ CH_{4}O}{32g\ CH_{4}O}) = 1.875mol\ CH_{4}O[/tex]

There is a higher number of moles of CH4O than C3H6O2, so the C3H6O2 will run out faster.  Therefore, C3H6O2 is the limiting reactant.
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