4.29 moles of NH3 -->? grams

How many grams of NH3 can be produced from 4.29 mol of N2 and excess H2. Please explain how you got the answer.

This is a stoichiometry problem, and the first thing you need is a balanced equation. I generally write the equation, balance it, then write the quantities (those given and asked for ( ?) ) below the substances, and write the the molar masses (from the periodic table) for any substances that must be converted between mass and mol.

   17.02 g/mol        (molar masses)

      N2         +      3 H2  →2 NH3             (balanced equation)

   4.29 mol             xs               ? g               (quantities given, asked)

To do stoichiometry, you will use the mol ratio of two substances in the balanced equation. We can only relate the mol of reactants to the mol of any other substance in the equation. We measure amounts of substances by mass (g) in the lab, however, so amounts are often given or asked for by mass, rather than mol. In this problem, you will have to convert mol NH3 to g NH3.

2 steps: 1)  Use the mol ratio from the balanced eq'n to calculate mol NH3 produced (stoichiometry)

           2)  Convert the mol NH3 to mass NH3 (in g) using dimensional analysis

1) The mol ratio from the balanced eq'n that you need is mol substance you are asked about

over the mol substance you are given

From the balanced eq'n: 2 mol NH3/1 mol N2

4.29 mol N2   x  2 mol NH3/1 mol N2  =  8.58 mol NH3    (stoichiometry)

2) 8.58 mol NH3 x   17.02 g NH3/1 mol NH3    =  146 g NH3      (conversion, mol to mass)

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Explanation:

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.