(a) Express the balanced chemical equation $$ \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} $$ in words, using the word mole(s) wherever appropriate. (b) To produce 1 mole of \(\mathrm{CO}_{2}\) from this reaction, how many grams of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) must you combine? (c) What is the theoretical yield in grams of \(\mathrm{H}_{2} \mathrm{O}\) for this reaction? (d) If you recover \(30.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), what is the percent yield? Answer: (a) One mole of methane molecules and 2 moles of oxygen molecules react to give 2 moles of water molecules and 1 mole of carbon dioxide molecules. (b) The molar mass of methane is \((12.011 \mathrm{~g} / \mathrm{mol} \mathrm{C})+(4 \times 1.0079 \mathrm{~g} / \mathrm{mol} \mathrm{H})=\) \(=16.043 \mathrm{~g} / \mathrm{mol} \mathrm{CH}_{4}\). This is the mass of \(1 \mathrm{~mole}\) of methane, which is what the reaction calls for. The molar mass of oxygen, \(\mathrm{O}_{2}\), is \((2 \times 15.999 \mathrm{~g} / \mathrm{mol} \mathrm{O})=31.998 \mathrm{~g} / \mathrm{mol}\). This is the mass of 1 mole of \(\mathrm{O}_{2}\). Because the reaction calls for 2 moles, we multiply by 2 to get \(63.996 \mathrm{~g}\) of \(\mathrm{O}_{2}\) needed. (c) The most we can hope to form is 2 moles of water. The molar mass of water is \((2 \times 1.0079 \mathrm{~g} / \mathrm{mol} \mathrm{H})+(15.999 \mathrm{~g} / \mathrm{mol} \mathrm{O})=18.015 \mathrm{~g} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O} .\) This is the mass of 1 mole of water. So the theoretical yield is just twice this, which is \(36.030 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). (d) \% yield \(=(30.0 \mathrm{~g} / 36.030 \mathrm{~g}) \times 100=83.3 \%\)

Step by step solution

01

(a) Chemical equation in words

The given balanced chemical equation is: \[ \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \] In words, this means: one mole of methane (CH₄) reacts with two moles of oxygen (O₂) to produce two moles of water (H₂O) and one mole of carbon dioxide (CO₂).

02

(b) Masses of CH₄ and O₂ required

First, calculate the molar masses of CH₄ and O₂: Molar mass of CH₄ = (12.011 g/mol C) + (4 × 1.0079 g/mol H) = 16.043 g/mol CH₄ Molar mass of O₂ = 2 × 15.999 g/mol O = 31.998 g/mol O₂ To produce 1 mole of CO₂, we need 1 mole of CH₄ and 2 moles of O₂. Therefore, required masses are: CH₄: 1 mole × 16.043 g/mol = 16.043 g O₂: 2 moles × 31.998 g/mol = 63.996 g

03

(c) Theoretical yield of H₂O

The theoretical yield of H₂O is the maximum amount of product formed under ideal conditions. According to the balanced chemical equation, 2 moles of H₂O are formed. Calculate the molar mass of H₂O: Molar mass of H₂O = (2 × 1.0079 g/mol H) + (15.999 g/mol O) = 18.015 g/mol H₂O Thus, theoretical yield of H₂O = 2 moles × 18.015 g/mol = 36.030 g

04

(d) Percent yield

Actual recovered mass of H₂O is given as 30.0 g. Now, calculate the percent yield as: Percent yield = (Actual yield / Theoretical yield) × 100 Percent yield = (30.0 g / 36.030 g) × 100 = 83.3 %

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