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katrin2010 [14]
3 years ago
5

When 12 moles of o2 react with 1.1 mole of c10h8 what is the limiting reactant?

Chemistry
2 answers:
Trava [24]3 years ago
6 0
1.1   moles   of C10H8  is the  limiting  reagent  in   the  reaction between   reaction  C10H8   and   O2
.
C10H8  +  12O2  ---->  10CO2   + 4H2O

C10H8  is  the  limiting   reagent   since   1.1  moles  of  C10H8  is  totally   consumed   during  the  reaction
mariarad [96]3 years ago
6 0

Answer: C10H8 is the limiting Reactant

Explanation:

Ok, before delving into this question let us take some definitions.

WHAT IS A LIMITING REACTANT?

A limiting Reactant is also called a limiting reagent. From the English word ''limiting' that is setting a limit, a boundary to something. Thus, in Chemistry, a limiting reagent can be defined as the chemical substance (in a chemical reaction) that determines how much a chemical reaction can go on to generate product(s).

What is an Excess Reactant: an Excess Reactant is the excessive reactants in any chemical equation. Let just say, it is the opposite of a limiting reagent.

With all these been said, let us write down the balanced chemical equation of the reaction;

C10H8 + 12 O2 --------> 10 CO2 + 4H2O.

From the Chemical reaction above, we can see that one mole of C10H8 reacts with 10 moles of Oxygen,O2 to give 10 moles of Carbondioxide, CO2 and 4 moles of water,H2O.

After using 1 mole From C10H8 the reaction should likely come to an end. If one mole of C10H8 reacts completely with one mole of O2, means that there will be an EXCESS OF 11 moles. Thus, the LIMITING REAGENT/REACTANT IS C10H8.

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You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b
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Answer:

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Explanation:

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Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

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Answer:

The correct answer is - 29.45 / 100 x 25.6 = 7.5392 grams

Explanation:

It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:

mass of Ca = total amount of sample*percentage of calcium in sample /100

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M of Ca = 7.5392 grams

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