a. RFM CuCO₃ = 123.5
RFM CuO = 79.5
b. 159 g
c. 93.84%
Further explanation
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%
[tex]\tt %yield=\dfrac{149.2}{159}\times 100\%=93.8\%[/tex]%yield = (149.2/159) x 100% = 93.84%
