In a chemical reaction, 247 g of copper carbonate was heated and 149.2 g of copper oxide was made.

a. RFM CuCO₃ = 123.5

RFM CuO = 79.5

b. 159 g

c. 93.84%

Given

247 g of copper carbonate

149.2 g of copper oxide

Required

a. RFM(relative formula mass)

b. the mass of CuO

c. % yield

Solution

Reaction

CuCO₃⇒CuO+CO₂

a.

RFM CuCO₃ = 63.5+12+3.16=123.5

RFM CuO = 63.5 + 16 = 79.5

b. mol CuCO₃ :

mol = mass : MW

mol = 247 : 123.5

mol = 2

From equation, mol ratio CuCO₃ : CuO = 1 :1 , so mol CuO = 2

mass CuO = 2 x 79.5 = 159 g

c. % yield = (actual/theoretical) x 100%

%yield = (149.2/159) x 100% = 93.84%

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

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