Question

Given the balanced chemical equation below, how many grams of oxygen are needed to produce 221.3 grams H2O? Report your answer to the hundredths place.C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Answer 1

Answer: 392.71 g of O2 would be necessary to produce 221.3g of H2O

Step-by-step explanation:

The question requires us to calculate the mass of oxygen gas (O2) necessary to produce 221.3g of H2O, given the following balanced equation:

`C_6H_(12)O_6+6O_2\rightarrow6CO_2+6H_2O`

The problem presented can be classified as a mass to mass stoichiometry problem. To solve it, we'll need to follow the steps:

mass of H2O → moles of H2O → moles of O2 → mass of O2

First, let's determine the number of moles of H2O contained in 221.3g of it, knowing that the molar mass of H2O is 18.02g/mol:

`n=(m)/(MM)\rightarrow n_(H_2O)=(221.3g)/(18.02g/mol)=12.28mol`

(where n is the number of moles, m is the mass of the sample and MM corresponds to the molar mass of the compound)

Therefore, 12.28 moles of H2O need to be produced in the reaction.

Next, let's determine the number of moles of O2 necessary to produce 12.28 moles of H2O. According to the balanced chemical equation, 6 moles of O2 are necessary to produce 6 moles of H2O:

6 mol H2O ------------------- 6 mol O2

12.28 mol H2O ------------- x

Solving for x, we have that 12.28 moles of O2 would be necessary to produce 12.28 moles of H2O.

At last, we need to convert the calculated number of moles of O2 to its correspondent mass, in grams. Knowing that the molar mass of O2 is 31.98g/mol, we can calculate the mass of O2 as:

`\begin{gathered} n=(m)/(MM)\rightarrow m=n* MM \\ \\ m_(O_2)=12.28mol*31.98g/mol=392.71g \end{gathered}`

Therefore, 392.71 g of O2 would be necessary to produce 221.3g of H2O.