Challenge Photosynthesis reactions in green plants use...

Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ and oxygen. A plant has 88.0 $\mathrm{g}$ of carbon dioxide and 64.0 $\mathrm{g}$ of water available for photosynthesis. \begin{equation} \begin{array}{l}{\text { a. Write the balanced chemical equation for the reaction. }} \\ {\text { b. Determine the limiting reactant. }} \\ {\text { c. Determine the excess reactant. }} \\ {\text { d. Determine the mass in excess. }} \\ {\text { e. Determine the mass of glucose produced. }}\end{array} \end{equation}

Challenge Photosynthesis reactions in green plants use carbon dioxide and water
to produce glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ and oxygen. A plant has 88.0 $\mathrm{g}$ of carbon dioxide and
64.0 $\mathrm{g}$ of water available for photosynthesis.
\begin{equation}
\begin{array}{l}{\text { a. Write the balanced chemical equation for the reaction. }} \\ {\text { b. Determine the limiting reactant. }} \\ {\text { c. Determine the excess reactant. }} \\ {\text { d. Determine the mass in excess. }} \\ {\text { e. Determine the mass of glucose produced. }}\end{array}
\end{equation}

Key Concepts

Excess Reactant

The excess reactant is the substance that remains after the reaction has gone to completion because it is present in a greater amount than necessary to react with the limiting reactant. Calculating the amount of excess reactant involves subtracting what was actually consumed from the initial amount available.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed. It is identified by comparing the mole ratios of the reactants used in the reaction to those required by the balanced equation.

Theoretical Yield Calculation

The theoretical yield is the calculated amount of product expected to be formed in a chemical reaction based on the limiting reactant and the stoichiometry of the balanced equation. This concept is essential for predicting the efficiency of a reaction and planning quantitative chemical experiments.

Balanced Chemical Equation

A balanced chemical equation is a representation of a chemical reaction in which the number of atoms of each element is the same on both sides of the equation. This ensures that the law of conservation of mass is obeyed, meaning no atoms are lost or created during the reaction.

Stoichiometry

Stoichiometry involves using the quantitative relationships derived from a balanced chemical equation to calculate the amounts of reactants and products involved in a chemical reaction. This includes converting between mass and moles using molar masses, determining mole ratios between substances, and predicting yields.

Transcript

00:01 Okay, so to answer this question, first what we're going to do is answer question number a, or question a, which is writing the balanced chemical equation for the reaction.

00:14 So the chemical reaction for photosynthesis is co2 plus h2o, which gives you c6h12o6, which is your glucose, plus oxygen.

00:43 So to write the balanced equation, we're going to have to find the coefficients that coincide, that makes the left side equal to the right side in terms of carbons, oxygens, and hydrogens.

01:03 So we see that we have six carbons on the right and one carbon on the left.

01:10 So first thing we're going to do is multiply this by six, and once we do that, we know we have 12 oxygens, and 13 oxygens on the left, and then two plus six, so eight oxygens on the right.

01:30 To balance that out now, we're going to add six oxygens over here, and finally we're going to add six oxygens over here.

01:39 So now this should be a balanced chemical equation for photosynthesis.

01:46 Next, we're going to find what the limiting reactant is.

01:52 So to do this, in this question, we're given the mass of carbon dioxide and water.

01:59 The first step we're going to have to do is convert the mass to moles.

02:03 So let's start off with moles, which is denoted by n of co2, which is equal to your mass, which is 88 grams, over the molar mass of co2, which is 44 .01 grams per mole, and the moles for that is 1 .999, which you can round up to moles.

02:44 Next, we're going to find the moles for water, which is equal to mass over molar mass again, so 64 grams over 18 .01 grams per mole.

03:02 So you can cancel the grams like you can do over here, which will give us an answer of 3 .5.

03:17 So now what we're going to do is we're going to use the coefficients of the balanced equations to find out what the limiting reagents are.

03:31 So for this, we're going to do 2 divided by 6.

03:36 So we're going to do the moles of co2 divided by the coefficient, which is 6.

03:44 So 2 over 6 can be simplified to one third, right? and we're going to do the same over here.

03:51 I'm going to do 3 .5 over 6, which is approximately one half.

04:00 That means that since this integer over here, one third, is less than one half, we're going to denote that the limiting reactant is going to be co2, and the excess reactant is equal to h2o.

04:35 And that answers questions b and c.

04:39 Okay, now we're going to answer a question.

04:47 How about that? now we're going to move on to d, which is the mass in excess.

04:57 So to find out the mass in excess, we're going to be working with moles again...

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