Question

The balanced chemical equation for the complete combustion of C2H2 is:

2C2H2 + 5O2 → 4CO2 + 2 H2O

How many grams of O2 are required to react with exactly 0.500 mol of C2H2?

1.25 g
20.0 g
40.0 g
25.0 g

Answer 1

Hey there:

Given the reaction:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

2 mol C2H2 ---------------- 5 mol O2

0.500 mol ---------------- moles O2 ??

moles O2 = 0.500 * 5 / 2

moles O2 = 2.5 / 2

=> 1.25 moles of O2

Molar mass O2 is 31.99 g/mol , so:

1 mol O2 -------------- 31.99 g

1.25 moles --------- mass O2 ??

mass O2 = 1.25 * 31.99 / 1

mass O2 = 40.0 / 1

mass O2 = 40.0 g

Answer C

Hope that helps!

Answer 2

Option C

Step-by-step explanation:

This is a problem of Stechiometry, so, to do this, let's write the equation again:

2C2H2 + 5O2 → 4CO2 + 2 H2O

According to the reaction, we can see that 2 moles of C2H" reacts with 5 moles of O2. So if we have 0.5 moles of C2H2, we only need to calculate the moles of O2 according to the above relation.

If: moles C2H2/moles O2 = 2/5

Then: 0.5 / moles O2 = 2/5

Solving for moles O2:

2 moles O2 = 5 * 0.5

moles O2 = 5 * 0.5 / 2

moles O2 = 1.25 moles

Now that we have the moles, we can calculate the grams, using the molecular weight of O2, which is 32 g/mol, then the mass:

m = moles O2 * MM

Replacing we have:

m = 1.25 * 32

m = 40 g O2

And this is the mass required to react exactly with the 0.5 moles of C2H2.