how many grams of carbon dioxide can be produced when 15.g of oxygen is reacted with 25.g of c2h2
To determine the number of grams of carbon dioxide produced when 15.0 g of oxygen is reacted with 25.0 g of C2H2, we need to use the balanced chemical equation for the reaction between oxygen and C2H2.
The balanced chemical equation is:
2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O
From the equation, we can see that two moles of C2H2 react with five moles of O2 to produce four moles of CO2. Therefore, we need to convert the grams of C2H2 and O2 to moles, and then use the mole ratio to determine the grams of CO2 produced.
First, we calculate the number of moles of C2H2:
Molar mass of C2H2 = 12.01 g/mol (carbon) + 2 * 1.01 g/mol (hydrogen)
= 26.04 g/mol
Number of moles of C2H2 = mass of C2H2 / molar mass of C2H2
= 25.0 g / 26.04 g/mol
= 0.960 mol
Next, we calculate the number of moles of O2:
Molar mass of O2 = 16.00 g/mol (oxygen) * 2
= 32.00 g/mol
Number of moles of O2 = mass of O2 / molar mass of O2
= 15.0 g / 32.00 g/mol
= 0.469 mol
Now, we can compare the mole ratio between C2H2 and O2 in the balanced chemical equation:
C2H2 : O2 = 2 : 5
Since the ratio is 2:5, we need to find the limiting reagent (the reactant that is completely consumed). The reactant that produces fewer moles of CO2 is the limiting reagent.
By comparing the moles of C2H2 and O2, we can see that the mole ratio of C2H2 to O2 is:
0.960 mol C2H2 / 2 = 0.480 mol C2H2
0.469 mol O2 / 5 = 0.094 mol O2
Since 0.094 mol of O2 is the smaller amount, it is the limiting reagent.
Using the mole ratio from the balanced equation, we can determine the moles of CO2 produced:
C2H2 : CO2 = 2 : 4
Moles of CO2 = moles of O2 * (4/5)
= 0.094 mol * (4/5)
= 0.0752 mol
Finally, we can convert the moles of CO2 to grams:
Molar mass of CO2 = 12.01 g/mol (carbon) + 2 * 16.00 g/mol (oxygen)
= 44.01 g/mol
Grams of CO2 = moles of CO2 * molar mass of CO2
= 0.0752 mol * 44.01 g/mol
= 3.31 g
Therefore, when 15.0 g of oxygen is reacted with 25.0 g of C2H2, approximately 3.31 grams of CO2 will be produced.