Asked by Anonymous
Two more need help figuring out how to do:
How many liters of water vapor can be produced when 8.23 liters of oxygen gas reacts with hydrogen gas (H2g + 02g = H20g.
How many liters of water vapor can be produced if 108 grams of methane gas are conbusted at 312K and 98 atm. CH4 + 202 = C02 + 2H20
How many liters of water vapor can be produced when 8.23 liters of oxygen gas reacts with hydrogen gas (H2g + 02g = H20g.
How many liters of water vapor can be produced if 108 grams of methane gas are conbusted at 312K and 98 atm. CH4 + 202 = C02 + 2H20
Answers
Answered by
drwls
In your first qeustion, you need to state what the temperature and pressure of the oxygen is. Your chemical reaction equation is not balanced.
108 g of methane is 108/16 = 6.75 moles. The reaction will produce 13.5 moles of H2O. Convert that to liters using the perfect gas law.
108 g of methane is 108/16 = 6.75 moles. The reaction will produce 13.5 moles of H2O. Convert that to liters using the perfect gas law.
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