Question

A chemist burned 150.0 g of C3H8 in the presence of 300.0 g of O2 and obtained 210.43 g of CO2. What is the percent yields of the combustion reaction?

Answer 1

Solution.

$m(C_3H_8)=150.0g;$

$m(O_2)=300.0g;$

$m(CO_2)=210.43g;$

150.0g 300.0g 210.43g

$C_3H_8+5O_2\to 3CO_2+4H_2O;$

$\nu(C_3H_8)=\dfrac{150.0}{44}=3.41mol;$

$\nu(O_2)=\dfrac{300.0}{32}=9.375mol;$

3.41mol xmol

$C_3H_8+5O_2\to 3CO_2+4H_2O;$

1 5

$x=\dfrac{3.41 mol\sdot5}{1}=17.05;$

9.375mol y

$C_3H_8+5O_2\to 3CO_2+4H_2O;$

5 3

$\nu(CO_2)=\dfrac{9.375\sdot3}{5}=5.625mol;$

$m_t(CO_2)=5.625\sdot44=247.5g;$

$\eta=\dfrac{m(CO_2)}{m_t(CO_2)};$

$\eta=\dfrac{210.43}{247.5}=0.85=85\%;$

Answer: $\eta=85\%.$