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Step-by-Step Guide to Solving Stoichiometry Problems in Chemistry

This guide by Aryn Balogh walks you through the process of solving a stoichiometry problem using water decomposition as an example. Start with writing a balanced chemical equation for 12.7 grams of water decomposing into hydrogen and oxygen gases. Follow each step, from converting grams to moles, finding the moles of products using coefficients, to converting moles back into grams. Finally, verify the law of conservation of mass. This systematic approach ensures accurate calculations of produced gases.

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Step-by-Step Guide to Solving Stoichiometry Problems in Chemistry

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  1. Steps for Solving a Stoichiometry Problem By: Aryn Balogh Period: 1

  2. If 12.7g of water is decomposed using an electric current, how many grams of Hydrogen and Oxygen gas will be produced?

  3. Step 1:Write a complete and balanced chemical equation. 2H2O 2H2 + O2

  4. Step 2:Draw a column for each chemical. 2H2O 2H2 + O2

  5. Step 3: Write the amount given in the appropriate column. 2H2O 2H2 + O2 12.7g

  6. Step 4:Convert the amount given into moles. 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705 moles

  7. Step 5:Find moles for each of the other chemicals. a) In each of the other columns write the moles of given (x) a fraction. 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705 moles 0.705mx X/X 0.705mx X/X

  8. b) The numerator of the fraction is the coefficient of that column. 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705m x 2/X= 0.705m x 1/X=

  9. c) The denominator of the fraction is the coefficient of the given column. 2H2O 2H2 + O2 0.705m x 1/2= 12.7g/1 x 1mole/18.0148g= 0.705m x 2/2=

  10. d) Do math and label as moles. 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705 moles 0.705m x 2/2= 0.705 moles 0.705m x 1/2= 0.353 moles

  11. Step 6:Convert all moles into grams. 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705 moles 0.705m x 2/2= 0.705 moles 0.705m x 1/2= 0.353 moles 0.705m/1 x 2.0158g/1m= 1.42g 0.353m/1 x 31.998g/1m= 11.3g

  12. Step 7:Verify the law of conservation of mass 2H2O 2H2 + O2 12.7g/1 x 1mole/18.0148g= 0.705 moles 0.705m x 2/2= 0.705 moles 0.705m x 1/2= 0.353 moles 0.705m/1 x 2.0158g/1m= 1.42g 0.353m/1 x 31.998g/1m= 11.3g 12.7g = 1.42g + 11.3g 12.7g = 12.7g 1.42g of Hydrogen gas will be produced & 11.3g of Oxygen will be produced.

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