At 506 K and 2.05 atm, if 6.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3? (R = 0.08206 L atm/mol K)

Explanation:

You must convert the 6.00 grams of H₂ into number of moles, and then use the stoichiometry of the reaction to find the number of moles of NH₃, which can be converted into volume using the ideal gas equation.

1. Number of moles of H₂

2. Number of moles of NH₃

i) Chemical equation:

ii) Mole ratio:

iii) Proportion:

3. Volume of NH₃

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