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2. The heat from an acetylene torch is produced by burning acetylene (C2H2) in oxygen.

2C2H2 + 502 --> 4CO2 + 2H20

When 2.40mol C2H2 reacts with 7.40mol O2,

a. Which reactant is the limiting reactant?

b. How many grams of water can be produced by the reaction?

c. How many moles of the excess reactant remains?



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Sagot :

Answer: a. C₂H₂

              b. 43.2 g

              c. 1.4 moles

Explanation: In a chemical reaction, when a reagent is completely consumed and limits the product formed is called Limiting Reagent. The reactant that is left is the Excess Reactant.

For the burning of acetylene:

[tex]2C_{2}H_{2}+5O_{2} \rightarrow 4CO_{2}+2H_{2}O[/tex]

a. An easy way to determine which reactant is limiting, is to divide the number of moles of each reactant by the coefficient in the balanced reaction. The component that gives the smallest number is the limiting reagent.

The ratio for acetylene is

[tex]\frac{2.4}{2}[/tex] = 1.2

For oxygen:

[tex]\frac{7.4}{5}=[/tex] 1.48

So, the limiting reactant is acetylene.

b. From the balanced reaction, for each 2 moles of acetylene is consumed, 2 moles of water is produced. Then, for 2.4 moles, it will be produced 2.4 moles of water.

Mol is mass in grams divided by molar mass of the component.

[tex]n=\frac{m}{M}[/tex]

Molar mass of water is M = 18 g/mol.

Then, mass of water produced:

m = n.M

m = 2.4(18)

m = 43.2 g

With 2.4 moles of acetylene, it will be produced 43.2 grams of water.

c. For each 2 moles of acetylene consumed, 5 moles of oxygen is also consumed. So, for 2.4 moles:

2 moles = 5 moles

2.4 moles = n

[tex]n=\frac{2.4*5}{2}[/tex]

n = 6 moles

For 2.4 moles of acetylene, it is consumed 6 moles of oxygen. Then, the excess is

7.4 - 6 = 1.4

The excess reactant remaining is 1.4 moles of oxygen.

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