Gas stoichiometry at stp
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Gas Stoichiometry At STP. Start with a balanced equation . Convert mass to moles (use molar mass) or volume to moles (use 22.4 L/mol) , if necessary. Use mole ratios to calculate moles of unknown. Convert moles to mass (use molar mass) or volume (use 22.4 L/mol) , if necessary.

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Gas Stoichiometry At STP

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Gas stoichiometry at stp

Gas Stoichiometry At STP

  • Start with a balanced equation.

  • Convert mass to moles (use molar mass) or volume to moles (use 22.4 L/mol), if necessary.

  • Use mole ratios to calculate moles of unknown.

  • Convert moles to mass (use molar mass) or volume (use 22.4 L/mol), if necessary.


Gas stoichiometry

Gas Stoichiometry

If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.

3 H2(g) + N2(g)  2NH3(g)

3moles H2 +1mole N2 2moles NH3

3liters H2 + 1liter N2 2liters NH3


Volume volume calculation

Volume-Volume Calculation

How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?

3 H2(g) + N2(g)  2NH3(g)

12 L H2

2

L NH3

= L NH3

8.0

3

L H2

Volume ratio


Mass volume calculation

Mass-Volume Calculation

How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s)  2 KCl(s) + 3 O2(g)

Molar volume

50.0 g KClO3

1 mol KClO3

3 mol O2

22.4 L O2

122.55 g KClO3

2 mol KClO3

1 mol O2

= 13.7 L O2

Molar mass

Mole ratio


Gas stoichiometry not at stp

Gas Stoichiometry NOT at STP

  • Start with a balanced equation.

  • Convert to moles (if amount is L, use n = PV/RT).

  • Use mole ratios to calculate moles of unknown.

  • Convert moles to desired measurement (to convert to L use PV=nRT).


Mass volume not at stp

Mass-Volume NOT at STP

How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2KClO3(s)  2KCl(s) + 3O2(g)

50.0 g KClO3

1 mol KClO3

3mol O2

0.612

=

mol O2

122.55 g KClO3

2mol KClO3

= 16.7 L


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