Answer :
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.
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.
