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lsmartwork5-chemistry-answers: Stoichiometry Method to Get Correct Results

For lsmartwork5-chemistry-answers searches, how can a combustion stoichiometry problem be solved and checked correctly using a balanced equation, mole ratios, and unit cancellation?

Subject: General Chemistry Chapter: Chemical Reactions Topic: Stoichiometry of the Reaction Answer included
lsmartwork5-chemistry-answers chemistry answers Smartwork 5 chemistry stoichiometry dimensional analysis balanced chemical equation mole ratio combustion reaction
Accepted answer Answer included

Problem

The search phrase lsmartwork5-chemistry-answers often appears when a stoichiometry result does not match an expected submission format. A reliable way to obtain correct chemistry answers is to use a balanced chemical equation, dimensional analysis, and careful unit cancellation.

Proprietary answer keys for paid homework systems should not be used or redistributed. Correct results can be produced by a transparent calculation that shows (1) a balanced equation, (2) mole-to-mole conversion, (3) unit cancellation, and (4) significant-figure consistency.

Worked example (original problem)

Propane combusts completely in excess oxygen. A sample of 12.0 g of propane reacts. Determine the mass of carbon dioxide produced.

Step 1: Write and balance the combustion reaction

\[ \mathrm{C_3H_8 + 5\ O_2 \rightarrow 3\ CO_2 + 4\ H_2O} \]

Step 2: Convert the given mass of propane to moles

Use molar masses (g/mol): \(M(\mathrm{C_3H_8}) = 3 \cdot 12.01 + 8 \cdot 1.008 = 44.094\).

\[ n(\mathrm{C_3H_8}) = 12.0\ \text{g} \times \frac{1\ \text{mol}}{44.094\ \text{g}} = 0.272\ \text{mol} \]

Step 3: Use the mole ratio to find moles of CO2

From the balanced equation, \(1\ \text{mol}\ \mathrm{C_3H_8}\) produces \(3\ \text{mol}\ \mathrm{CO_2}\).

\[ n(\mathrm{CO_2}) = 0.272\ \text{mol}\ \mathrm{C_3H_8} \times \frac{3\ \text{mol}\ \mathrm{CO_2}}{1\ \text{mol}\ \mathrm{C_3H_8}} = 0.816\ \text{mol}\ \mathrm{CO_2} \]

Step 4: Convert moles of CO2 to grams

\(M(\mathrm{CO_2}) = 12.01 + 2 \cdot 16.00 = 44.01\ \text{g/mol}\).

\[ m(\mathrm{CO_2}) = 0.816\ \text{mol} \times \frac{44.01\ \text{g}}{1\ \text{mol}} = 35.9\ \text{g} \]

The given mass (12.0 g) has 3 significant figures, so the result is reported as 35.9 g CO2.

Compact calculation table

Stage Conversion Result Unit check
Mass → moles (propane) \(12.0\ \text{g} \times \frac{1\ \text{mol}}{44.094\ \text{g}}\) \(0.272\ \text{mol}\ \mathrm{C_3H_8}\) g cancels → mol
Mole ratio (to CO2) \(\times \frac{3\ \text{mol}\ \mathrm{CO_2}}{1\ \text{mol}\ \mathrm{C_3H_8}}\) \(0.816\ \text{mol}\ \mathrm{CO_2}\) mol C3H8 cancels
Moles → mass (CO2) \(\times \frac{44.01\ \text{g}}{1\ \text{mol}}\) \(35.9\ \text{g}\ \mathrm{CO_2}\) mol cancels → g

Visualization: stoichiometry “road map” for chemistry answers

Given quantity usually grams, liters, or solution molarity Moles (pivot unit) convert using molar mass or \(PV = nRT\), or \(M = \frac{n}{V}\) Requested quantity grams, moles, volume, or concentration use conversion factor units cancel apply mole ratio from balanced equation Final checks: (1) balanced equation, (2) units cancel to target unit, (3) significant figures match the limiting data, (4) magnitude is chemically reasonable.
Correct stoichiometry follows a consistent pathway: convert the given quantity to moles, use the balanced-equation mole ratio, then convert to the requested unit and verify unit cancellation.

Practical checklist for fixing a “wrong answer” submission

  • Equation first: coefficients must be balanced before any mole ratio is used.
  • Conversion factors written explicitly: each step should show a fraction so units cancel cleanly.
  • Molar masses consistent: use the same atomic masses across a problem set when possible (common rounding differences can shift the last digit).
  • Significant figures: the reported value should match the precision of the least precise measured input.
  • Sanity check: combustion of a hydrocarbon produces substantial CO2; a result far smaller than the starting mass is often a unit or ratio error.

Final result for the example

From 12.0 g of propane reacting completely in excess oxygen, the predicted product mass is 35.9 g CO2.

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