stoichiometry and gasses

Question

2) how many grams of water will need to be decomposed to make the partial pressure of hydrogen be 98.8 kPa in a 355 ml container at 22.5° c?

3) a 3.2 L container of nitrogen and hydrogen is at a temperature of 288k and allow to react to for ammonia (NH3) if the partial pressure on nitrogen was originally 0.53 ATM and the partial pressure of hydrogen was originally 0.47 ATM 

A) how many miles of nitrogen do you have in container 

B) how many moles of hydrogen do you have in your container 

C) what is my limiting reactant

D) how many grams of (NH3) will be formed

Answer 1

The chemical reaction for the decomposition of water is as follows:

 

2H₂0  --> 2H₂ + O₂     We can calculate the required grams of water.

 

PV = nRT   -->  n = PV/RT

 

n = moles of Hydrogen = 98.8 kPA *(0.335L) * 1 atm/101.3kpa * mole-K/(0.082057 L-atm * 295.5k) = 0.01347 moles

 

Now do the stoichiometry:

 

0.01347 moles H2 * 2 moles water/ 1 mole H2 * 18 grams water/mole water = 0.485 grams of water.

 

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The ratio of hydrogen and Nitrogen to combine is 3 H₂ to one N₂, so hydrogen is the limiting reactant since the given amount is less than three times the partial pressure of nitrogen.

 

You can calculate the number of moles of each using the ideal gas equation.   n = PV/(RT).   

Having calculated the number of moles of each, and having identified the limiting reactant, calculating the final amounts of NH3  is easily done given the balanced equation.   The molecular mass of Ammonia is approximately 17g/mole.   

 

3H₂ + N₂ --> 2NH₃