When 24.9 grams of ethane (c2h6) react with excess oxygen to produce carbon dioxide, the percent yield is 56.8%. what was the actual yield of the reaction? 2c2h6 + 7o2 4co2 + 6h2o

actual yield= 41.4 g

Explanation:

Reaction hydrocarbons with excess Oxygen is called Combustion reaction.

To find the actual yield, we need to know thereotical yield.

Theoretical yield:

(24.9 g C2H6/30 g/mol C2H6)X(4 molCO2/2 mol C2H6)X(44 g CO2/1 mol CO2)

= 73.04 g Co2

%yield= observed yiled x100/theoretical yield

Observed yield= %yieldxtheoretical yield/100

= 56.8x73.04/100

= 41.4 g

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol