Consider the reaction of methane with ammonia and oxygen.

2CH4 (g) + 2NH3 (g) + 3O2 (g) 2HCN (g) + 6H2O (l)

Determine the limiting reactant in a mixture containing 123 g of CH4, 114 g of NH3, and 423 g of O2. Calculate the maximum mass (in grams) of hydrogen cyanide, HCN, that can be produced in the reaction.

The limiting reactant is:

NH3

CH4

O2

Amount of HCN formed =

g

Limit reactant is NH3

181.0593 g of HCN

Explanation:

2CH4 (g) + 2NH3 (g) + 3O2 (g) 2HCN (g) + 6H2O (l)

weights:

mwCH4 = 16 g/mol

mwNH3 = 17 g/mol

mwO2 = 32 g/mol

mwHCN = 27 g/mol

mwH2O = 18 g/mol

actual reactants:

gCH4 = 123 g

gNH3 = 114 g

gO2 = 423 g

lets see hoy many moles are present for each reactive;

molCH4 = gCH4/mwCH4 = > 7.6875 mol

molNH3 = gNH3/mwNH3 = > 6.7059 mol

molO2 = gO2/mwNH3 = > 24.8824 mol

The ratio of r moles needed in this reaction:

CH4=molCH4/2mol = > 3.8438

NH3=molNH3/2mol = > 3.3529

O2=molO2/3mol = > 8.2941

The smallest ratio is NH3, then this is the limit reactant.

then we can tell how much HCN is going to be produced:

if 2 moles of NH3 produce 2 Moles of HCN, then 6.7059 mol of NH3 will produce 6.7059 mol of HCN

gHCN = 6.7059 mol*mwHCN = > 181.0593 g

HCN