Question

1. complete the table below for the reaction of methane:

1. what is always the simplified ratio of 2 ch₄ used: 4 o₂ used: 2 co₂ made: 4 h₂o made?

this means that for every 1 mole of ch₄ used, you will use __ times as much o₂ and will make times as much co₂ and __ times as much h₂o

1. predict: (show/explain your process for each)

a. if you used 9 ch₄, how many o₂ would you use?
9 ch₄ x ______ = ______ o₂ used
b. if you used 9 ch₄, how many co₂ would you make?
9 ch₄ x ______ = ______ co₂ made

1. what happens if the amounts of ch₄ and o₂ dont match the ratio listed above? describe what happened in the last scenario in your own words. why did this happen? explain.

Explanation:

Step1: Simplify the ratio

The ratio \(2:4:2:4\) can be simplified by dividing each number by \(2\).
\(2\div2 = 1\), \(4\div2=2\), \(2\div2 = 1\), \(4\div2=2\). So the simplified ratio is \(1:2:1:2\).
This means for every \(1\) mole of \(CH_{4}\) used, you will use \(2\) times as much \(O_{2}\) (\(1\times2\)), make \(1\) times as much \(CO_{2}\) (\(1\times1\)) and \(2\) times as much \(H_{2}O\) (\(1\times2\)).

Step2: Predict \(O_{2}\) used for \(9CH_{4}\)

From the ratio \(CH_{4}:O_{2}=1:2\). If we have \(9CH_{4}\), then the amount of \(O_{2}\) used is \(9\times2\).

Step3: Predict \(CO_{2}\) made for \(9CH_{4}\)

From the ratio \(CH_{4}:CO_{2}=1:1\). If we have \(9CH_{4}\), then the amount of \(CO_{2}\) made is \(9\times1\).

Step4: Explain leftovers

In a chemical reaction, reactants react in a fixed ratio (from the balanced chemical equation \(CH_{4}+2O_{2}=CO_{2}+2H_{2}O\)). If the amounts of \(CH_{4}\) and \(O_{2}\) don't match the ratio (\(1:2\) in the balanced equation), one of the reactants will be in excess (leftover). In the last scenario (\(4CH_{4}\) and \(5O_{2}\)), based on the ratio \(CH_{4}:O_{2}=1:2\), for \(4CH_{4}\) we need \(4\times2 = 8O_{2}\), but we have only \(5O_{2}\). \(O_{2}\) is the limiting reactant. \(2CH_{4}\) will react (\(2CH_{4}+4O_{2}=2CO_{2}+4H_{2}O\)), leaving \(4 - 2=2CH_{4}\) and \(5 - 4 = 1O_{2}\) (but actually, based on the stoichiometry, \(O_{2}\) is consumed as per its amount, and \(CH_{4}\) is in excess because \(O_{2}\) runs out first).

Answer:

1. Simplified ratio: \(1:2:1:2\). For every \(1\) mole of \(CH_{4}\) used, \(2\) times \(O_{2}\), \(1\) time \(CO_{2}\), \(2\) times \(H_{2}O\).
2. a. \(9CH_{4}\times2 = 18O_{2}\) used. b. \(9CH_{4}\times1=9CO_{2}\) made.
3. If amounts don't match the ratio (from balanced equation \(CH_{4}+2O_{2}=CO_{2}+2H_{2}O\)), one reactant is leftover. In the last scenario (\(4CH_{4}\), \(5O_{2}\)), \(O_{2}\) is limiting. \(2CH_{4}\) react (\(2CH_{4}+4O_{2}=2CO_{2}+4H_{2}O\)), leaving \(2CH_{4}\) (because \(O_{2}\) is consumed completely based on its amount and the stoichiometric ratio).