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%

i'm pretty sure it is letter c.) car