What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2? 4C3H9N+25O2=>12CO2+18H2O+4NO2

Answer 1

The limiting reactant would be O₂.

The balanced equation for the reaction is

4C₃H₉N + 25O₂ →12CO₂ + 18H₂O + 4NO₂

To determine the limiting reactant, we calculate the amount of product that can be formed from each of the reactants. Whichever reactant gives the smaller amount of product is the limiting reactant.

Let's use CO₂ as the product.

From C₃H₉N:

26.0 g C₃H₉N × #(1"mol C₃H₉N")/(59.11"g C₃H₉N") × (12"mol CO₂")/(4"mol C₃H₉N")# = 1.32 mol CO₂

From O₂:

46.3 g O₂ × #(1"mol O₂")/(32.00"g O₂") × (12"mol CO₂")/(25"mol O₂")# = 0.694 mol CO₂

The O₂ gives the smaller amount of CO₂, so the O₂ is the limiting reactant.

Hope this helps.

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Answer 2

Cooper Anderson

To determine the limiting reagent, calculate the moles of each reactant. Then, use stoichiometry to find the amount of product formed from each reactant. The reactant that produces the least amount of product is the limiting reagent.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2? 4C3H9N+25O2=&gt;12CO2+18H2O+4NO2

What would be the limiting reagent if 26.0 grams of C3H9N were reacted with 46.3 grams of O2? 4C3H9N+25O2=>12CO2+18H2O+4NO2