In the combustion of propane, $C_3H_8$, 32 g of $O_2$ is needed to form 10 g of $H_2O$. What are the theoretical yield and percentage yield of $H_2O$? ($C = 12, O = 16, H = 1$)
$C_3H_8(g) + 5O_2(g) \longrightarrow 3CO_2(g) + 4H_2O(1)$
Answer (2)
Calculate the moles of O 2 used: n ( O 2 ) = 32 32 = 1 mole.
Determine the theoretical moles of H 2 O produced: n t h eore t i c a l ( H 2 O ) = 5 4 × 1 = 0.8 moles.
Calculate the theoretical yield of H 2 O : ma s s t h eore t i c a l ( H 2 O ) = 0.8 × 18 = 14.4 g.
Calculate the percentage yield of H 2 O : P erce n t a g e Yi e l d = 14.4 10 × 100 = 69.44% . The theoretical yield is 14.4 g and the percentage yield is 69.44% .
Explanation
Problem Setup and Given Information We are given the balanced chemical equation for the combustion of propane: C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 C O 2 ( g ) + 4 H 2 O ( l ) We are also given that 32 g of O 2 is used to produce 10 g of H 2 O . We need to find the theoretical yield and percentage yield of H 2 O .
Calculating Moles of Oxygen First, we need to calculate the number of moles of O 2 used. The molar mass of O 2 is 2 × 16 = 32 g/mol. Therefore, the number of moles of O 2 is: n ( O 2 ) = M o l a r M a ss ( O 2 ) ma ss ( O 2 ) = 32 g / m o l 32 g = 1 m o l
Theoretical Moles of Water From the balanced equation, 5 moles of O 2 produce 4 moles of H 2 O . So, 1 mole of O 2 will produce 5 4 moles of H 2 O . Therefore, the theoretical number of moles of H 2 O produced is: n t h eore t i c a l ( H 2 O ) = 5 4 × n ( O 2 ) = 5 4 × 1 m o l = 0.8 m o l
Theoretical Yield of Water Next, we calculate the theoretical mass of H 2 O (theoretical yield). The molar mass of H 2 O is 2 × 1 + 16 = 18 g/mol. Therefore, the theoretical mass of H 2 O is: ma s s t h eore t i c a l ( H 2 O ) = n t h eore t i c a l ( H 2 O ) × M o l a r M a ss ( H 2 O ) = 0.8 m o l × 18 g / m o l = 14.4 g
Percentage Yield of Water Now, we calculate the percentage yield of H 2 O . The actual yield is given as 10 g. Therefore, the percentage yield is: P erce n t a g e Yi e l d = T h eore t i c a l Yi e l d A c t u a l Yi e l d × 100 = 14.4 g 10 g × 100 = 69.44% Therefore, the theoretical yield of H 2 O is 14.4 g and the percentage yield is approximately 69.44%.
Final Answer In summary, the theoretical yield of water is 14.4 g, and the percentage yield is approximately 69.44%.
Examples
In the field of chemical engineering, calculating theoretical and percentage yields is crucial for optimizing chemical reactions. For example, in the production of fertilizers, understanding the theoretical yield of ammonia ( N H 3 ) from the Haber-Bosch process helps engineers determine the maximum amount of ammonia that can be produced under ideal conditions. The percentage yield then indicates the efficiency of the actual production process, guiding adjustments to improve yield and reduce waste. This ensures cost-effectiveness and sustainability in industrial chemical processes.
The theoretical yield of water in the combustion of propane is 14.4 g, and the percentage yield is approximately 69.44%.
;
