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

the answer is c. nacl