A mixture of C3H8 and C2H2 has a mass of 2.5 g . It is burned in excess O2 to form a mixture of water and carbon dioxide that contains 1.5 times as many moles of CO2 as of water. Find the mass of C2H2 in the original mixture.
To solve this problem, we can follow these steps:
Step 1: Determine the moles of CO2 and water formed.
Given that the mole ratio of CO2 to water is 1.5, let's assume that "x" moles of water are formed. Therefore, the number of moles of CO2 formed is 1.5x.
Step 2: Write balanced equations for the combustion reactions of C3H8 and C2H2.
To do this, we need to know the balanced equations for the combustion reactions of propane (C3H8) and acetylene (C2H2):
C3H8 + 5O2 → 3CO2 + 4H2O
C2H2 + 2.5O2 → 2CO2 + H2O
From these equations, we can see that for every mol of C3H8 burned, we obtain 3 mol of CO2 and 4 mol of H2O, whereas for every mol of C2H2 burned, we get 2 mol of CO2 and 1 mol of H2O.
Step 3: Determine the moles of C3H8 and C2H2 in the original mixture.
Let's assume that "a" moles of C3H8 and "b" moles of C2H2 are present in the original mixture.
From the balanced equations, we can set up two equations based on the elements carbon (C) and hydrogens (H):
For carbon: 3a + 2b = moles of CO2 formed
For hydrogen: 4a + b = moles of H2O formed
Step 4: Solve the system of equations.
Since we know that there are 1.5 times as many moles of CO2 as of water, we can set up the following equation:
1.5x = a = moles of CO2 formed
Substituting the value of "a" in the equation for carbon:
3(1.5x) + 2b = x
Expanding this equation:
4.5x + 2b = x
Rearranging the equation:
2b = x - 4.5x
2b = -3.5x
Dividing by 2:
b = -1.75x
Step 5: Calculate the moles of water formed.
Using the equation for hydrogen:
4a + (-1.75x) = x
Simplifying the equation:
4a - 1.75x - x = 0
4a - 2:75x = 0
Step 6: Calculate the mass of C2H2 in the mixture.
Since we know the mass of the mixture is 2.5 g, we can use the molar masses of C2H2 (26 g/mol) and C3H8 (44 g/mol) to calculate the moles of C2H2 and C3H8.
Let's assume the molar mass of C3H8 = y g.
a(molar mass of C3H8) + b(molar mass of C2H2) = mass of mixture
ay + (1.75x)(26 g/mol) = 2.5 g
Substituting the value of b from Step 4:
ay + (-1.75x)(26 g/mol) = 2.5 g
Simplifying the equation:
ay - 45.5x = 2.5 g
Since we have two unknowns, we need another equation to solve this system of equations.
To solve this problem, we need to use stoichiometry and mole ratios to find the mass of C2H2 in the original mixture.
Let's proceed step by step:
1. Determine the balanced equation for the combustion reaction between C3H8 and O2. The balanced equation is:
C3H8 + 5O2 -> 3CO2 + 4H2O
2. Calculate the moles of CO2 and H2O formed from the given mass of the mixture.
The molar mass of CO2 is 44 g/mol and the molar mass of H2O is 18 g/mol.
Moles of CO2 = (1.5 x Moles of H2O) // Given in the problem.
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= Moles of CO2 x 44 g/mol
Mass of H2O = Moles of H2O x Molar mass of H2O
= Moles of H2O x 18 g/mol
3. Calculate the moles of C3H8 burned in the combustion reaction.
From the balanced equation, we know that the mole ratio between C3H8 and CO2 is 1:3.
Moles of C3H8 = (1/3) x Moles of CO2
= (1/3) x (Mass of CO2 / Molar mass of CO2)
4. Calculate the moles of C2H2 in the original mixture.
From the balanced equation, we know that the mole ratio between C3H8 and C2H2 is 1:1.
Moles of C2H2 = Moles of C3H8
= (1/3) x (Mass of CO2 / Molar mass of CO2)
5. Finally, calculate the mass of C2H2 in the original mixture.
Mass of C2H2 = Moles of C2H2 x Molar mass of C2H2
By following these steps, you should be able to calculate the mass of C2H2 in the original mixture.
Two equations to be solved simultaneously.
Let X = g C3H8
and Y = g C2H2
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X + Y = 2.5 (equation 1)
mols CO2 from C3H8 + mols CO2 from C2H2 = 1.5*(mols H2O from C3H8 + mols H2O from C2H2). Turning that into a second equation
(4X/44) + (3X/44) = 1.5[(Y/26) + (2Y/26)] (equation 2)
Solve for Y.
Post your work if you get stuck.