PLS HELP When 50.0 g of copper was reacted with 300.0 g of silver nitrate solution, 149 g of silver was obtained. What is the percent yield of silver obtained?

Percent yield of silver obtained = 78.18% (Approx.)

Explanation:

Given:

Amount of copper = 50 gram

Amount of silver nitrate = 300 gram

Amount of silver = 149 gram

Find:

Percent yield of silver obtained

Computation:

Percentage of Silver in silver nitrate =  108[100/(108+14+48)]

Percentage of Silver in silver nitrate = 108/[100/170]

Percentage of Silver in silver nitrate = 63.53% (Approx.)

Amount of Silver produced = [63.53%][300]

Amount of Silver produced = 190.59 gram

Percent yield of silver obtained = [Amount of silver / Amount of silver produced]100

Percent yield of silver obtained = [149/190.59]100

Percent yield of silver obtained = 78.18% (Approx.)

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