Answer:
131.7 grams of carbon dioxide would be formed
Explanation:
2C₂H₂ + 5O₂ = 4CO₂ + 2H₂O
m(CO₂)/{4M(CO₂)} = m(C₂H₂)/{2M(C₂H₂)}
m(CO₂)=2M(CO₂)m(C₂H₂)/M(C₂H₂)
m(CO₂)=2*44g/mol*38.9g/26g/mol = 131.7 g
131.7 grams of carbon dioxide would be formed
Answer: The mass of carbon dioxide produced in the given reaction is 132 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of ethyne = 38.9 g
Molar mass of ethyne = 26 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ethyne}=\frac{38.9g}{26g/mol}=1.5mol[/tex]
[tex]2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of ethyne produces 4 moles of carbon dioxide.
So, 1.5 moles of ethyne will produce = [tex]\frac{4}{2}\times 1.5=3mol[/tex] of carbon dioxide.
Molar mass of carbon dioxide = 44 g/mol
Moles of carbon dioxide = 3 moles
Putting values in equation 1, we get:
[tex]3mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=132g[/tex]
Hence, the mass of carbon dioxide produced in the given reaction is 132 grams.
