Question

in the balance equation 2c2h6+7o2=4co2+6h2o 10 g of c2h6 react with 42.5 g o2 what is the excess reactant

Answer 1

Answer:

O₂ will be the excess reactant.

Explanation:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

The excess reagent is one of which you have more than is necessary for the chemical reaction to take place. Then the excess reagent is one that is not completely depleted during the reaction.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So

$x=\frac{c*b}{a}$

To know the limiting and excess reagent you must first know the mass per mole of each compound. For this you know that:

• C: 12 g/mol
• H: 1 g/mol
• O: 16 g/mol

Given that the subindices indicate the amount of each element present in each compound, the molar mass of each reagent can be calculated:

• C₂H₆: 2*12 g/mol + 6*1 g/mol= 30 g/mol
• O₂: 2*16 g/mol= 32 g/mol

The reaction stoichiometry of the reaction (that is, the ratio between the amount of reagents and products in a chemical reaction) indicates that 2 moles of C₂H₆ and 7 moles of O₂ react, so the reacting mass of each compound is:

• C₂H₆: 2*30 g/mol= 60 g/mol
• O₂: 7*32 g/mol= 224 g/mol

Then, knowing the stoichiometric mass and using the given mass of data, the limiting reagent can be calculated using a rule of three: if 60 g of C₂H₆ react 224 g of O₂, how much mass of C₂H₆ will react with 42.5 g of O₂?

$massofC_{2} H_{6} =\frac{42.5gramsofO2*60gramsofC2H6}{224gramsofO2}$

mass of C₂H₆= 11.38 grams

But 11.38 grams of C₂H₆ are not available, 10 grams are available. Since you have less mass than you need to react with 42.5 grams of O₂, C₂H₆ will be the limiting reagent. This means that O₂ will be the excess reactant.

Answer 2

Answer;

Ethane (C2H6) is the excess reactant

Explanation;

From the equation;

2C2H6 + 7 O2 → 4CO2 + 6 H2O

2 moles of ethane requires 7 moles of oxygen to produce 4 moles of CO2 and 6 moles of water.

The mole ratio of ethane : oxygen is 2 : 7

10 g of C2H6 = 10/30 g/mol = 0.333 moles

12.5 g of Oxygen = 42.5/32 g/mol = 1.328 moles

Using C2H6 as limiting; 0.333 moles pf C2H6 requires 0.333× 7 =2.331 moles of oxygen

Using oxygen as limiting; 1.328 moles of oxygen requires, 1.328/7 = 0.1897 moles of ethane.

Therefore; Ethane is the excess reactant by 0.333 - 0.1897 = 0.1433 moles