In a particular reaction between copper metal and silver nitrate, 12.7 g Cu produced 38.1 g Ag. What is the percent yield of silver in this reaction? Cu + 2 A g N O3→ Cu(NO3)2+ 2Ag

Explanation:

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88 %

Explanation:

This is question we can solve by using the stoichiometry for the balanced chemical equation:

Cu + 2 AgNO₃   ⇒  Cu(NO₃)₂  + 2 Ag

Given the mass of copper we can calculate how many moles it represents, and from the stoichiometry of the reaction we know 1 mol Cu produces 2 moles Ag, so we can calculate the moles Ag that theoretically should  have been produced.

Finally we can determine the number of  moles of  Ag actually produced and calculate the percent yield.

We will need the atomic weights of Cu and Ag   (63.54 g/mol, 107.87 g/mol respectively).

# mol Cu = 12.7 g x 1mol/63.54 g/mol = 0.20 mol

# mol theoretical Ag produced = (2 mol Ag/ 1 mol Cu) x 0.20 mol Cu  

                                                   = 0.40 mol Ag theoretical

# mol Ag actually produced =  38.1 g Ag / 107.87 g/mol = 0.35 mol

The percent yield is then:

= (0.35 / 0.40) x 100 = 88 %                                            

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