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How many molecules of C3H8 are required to react with 10 molecules of O2? C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H2O(l)

Answer 1

Caleb Adams

Two molecules of #"C"_3"H"_8# are required.

The equation that is balanced is

#"C"_3"H"_8 + "5O"_2 → "3CO"_2 + "4H"_2"O"#

This tells you that five molecules of #"O"_2# require one molecule of #"C"_3"H"_8#, so

Ten molecules of #"O"_2# require twice as many molecules of #"C"_3"H"_8# or two molecules.

This is one of those problems you could solve mentally.

However, this becomes more difficult with more complex numbers.

Next, we set up the calculation using a mole ratio.

A fraction created from the coefficients in the balanced equation is the mole ratio.

To convert the number of molecules or moles of one substance to the number of molecules or moles of another, we use the mole ratio.

As an illustration, in your equation

The mole ratio would be either #("1 molecule C"_3"H"_8)/("5 molecules O"_2)# or #("5 molecules O"_2)/("1 molecule C"_3"H"_8)#.

We would arrange the formula as

#10color(red)(cancel(color(black)("molecules O"_2))) × ("1 molecule C"_3"H"_8)/(5color(red)(cancel(color(black)("molecules O"_2)))) = "2 molecules C"_3"H"_8#

Take note of how the molar ratio is configured to provide the right units for the answer by causing the units to cancel.

We would have obtained the incorrect units if we had used the other ratio.

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Answer 2

Mateo Aguilar

According to the balanced chemical equation, 1 molecule of C3H8 reacts with 5 molecules of O2. Therefore, to react with 10 molecules of O2, you would need 2 molecules of C3H8.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How many molecules of C3H8 are required to react with 10 molecules of O2? C3H8(g) + 5 O2(g) ---&gt; 3 CO2(g) + 4 H2O(l)

How many molecules of C3H8 are required to react with 10 molecules of O2? C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H2O(l)