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50 grams of propane, C₃H₈, reacts with 120 grams of oxygen. What is the limiting reagent?

AC₃H₈
BO₂
CCO₂
DH₂O

HELP QUICK POINTS!!!! 50 grams of propane, C H , reacts with 120 grams of oxygen. What is the limiting reagent? A) C H B) O C) CO D) H O class=

Verified Answer :

im pretty sure it is a, but i may be wrong

igeclement43

Answer:Limiting reagent is

B) O₂

Explanation:

From the balanced reaction.

C3H8 + 5O2 ==> 3CO2 + 4H2O

there are two reagents(C3H8 and O2)and the limiting one is the one that is used up before the end of the reaction .

C3H8 : 02 = 1:5

Given mass 50g of propane

Molar mass = (3×12 + 8×1) =44g/mol

Mole= 50/44 = 1.136mol.

Given 120g of oxygen gas

Molar mass =32g/mol

Mole = 120/32= 3.75mol.

From the ratio, for the complete reaction, oxygen needs 5× of propane = 5×1.136= 5.68mol , but it has less (3.75mol)

Therefore O₂ is the limiting reagent.

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