If 15.0g c3h6, 10.0g o2, and 5.00g nh3 are sealed in a reactor, what mass of acrylonitrile can be produced if the reaction runs to 100% conversion?

2 C3H6 + 2 NH3 + 3 O2  -> 2 C3H3N + 6 H2O

Molar masses

C3H6: 42.08 g/mol
NH3: 17.03 g/mol
O2: 32 g/mol

CH3H3N: 53.06 g/mol
H2O: 18 g/mol

# of moles

C3H6: 15.0 g / [42.08 g/mol] = 0.356 mol
NH3: 5.00 g / [17.03 g/mol] =  0.294 mol
O2: 10.0 g / [32 g/mol] = 0.313 mol

Now wee need to find the limitant reactant

The theoretical proportion is 2:2:3, equivalent to 1:1:1.5

The proportion ot the given gases is 0.356 : 0.294 : 0.313, dividing by 0.294 that is equivalent to: 1.2 : 1 : 1.06, where you see that O2 is below the theoretical proportion and it is the limitant reactant.

Now, from the equation you have 3 mol of O2 produces 2 mol of acrylonitrile (C3H3N)

Then you will obtain 0.313 * 2/3 = 0.209 mol acrylonitrile

The mass is 0.209mol * 53.06 g/mol = 11.07 g acrylonitrile

11.07 g

30litre of li2o is 1.0002mole approximately 1 mole