When a mixture of 10 g of acetylene and 10 g of oxygen is ignited, the result is combustion .

Question

The balanced equation is 2 C2H2 5 O2 > 4 CO2 2 H2O. You have 10 g of C2H2 and 10 g of O2. Which is the limiting reactant? and How many grams of C2H2, O2, CO2, and H2O are present after the reaction is complete. I am pretty sure that oxygen is the limiting reactant but can't figure out the grams of the excess reactant and products. Please help. Thank you!

Answer 1

Moles C2H2 = 10 g / 26.038 g/mol = 0.384

moles O2 = 10 g / 31.9988 g/mol = 0.313

the ratio between C2H2 and O2 is 2 : 5 so O2 is the limiting reactant

The ratio between O2 and CO2 is 5 : 4

Moles CO2 = 0.313 x 4 / 5 = 0.250

Mass CO2 = 0.250 mol x 44.0098 g/mol =11.0g

The ratio between O2 and H2O is 5 : 2

Moles H2O = 0.313 x 2 / 5 = 0.125

Mass H2O = 0.125 mol x 18.02 g/mol = 2.26 g

The ratio between C2H4 and O2 is 2 : 5

Moles C2H4 needed = 0.313 x 2 / 5 = 0.125

Moles C2H4 in excess = 0.384 - 0.125 =  0.259

Mass C2H4 in excess = 0.259 mol x 26.038 g/mol =6.74 g

Answer 2

2*26g acetilene need 5*32g oxygen.so,O2 is limiting.after combustion there is none acetylene,it decomposes: C2H2 >2C + H2 (the excess).that.s from the fact that acetylene is "unstable"

Answer 3

Yes, oxygen is limiting.  To do any problem like this you first have to convert weight to moles, because each molecule weighs a different amount.   MW is 28 for acetylene and 32 for O2.  

You have 0.357 mole acetylene and 0.313 mole O2, not quite as much O2 as acetylene.

So if the reaction goes to completion (and in this case it probably will be an explosion) then when done you'll have:

zero moles of acetylene

.357-.313 moles of oxygen (the leftovers)     0.044 mole or 1.6 g

For the products, you will have the number of moles that comes from the consumed acetylene's .357 moles.  Since there are two in the equation, divide by 2 and .178 is the base quantity being multiplied:

CO2  .714 moles  31.4 g

H20   .357 moles  6.42 g