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nika2105 [10]
2 years ago
12

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

plenty of light, which reactant would be the limiting reactant?
Chemistry
1 answer:
alisha [4.7K]2 years ago
7 0

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

                                = 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

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Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

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