Chemistry, 24.01.2020 18:31 hurricane01

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

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Answer from: kamal81

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

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

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/molH: 1 g/molO: 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/molO₂: 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/molO₂: 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 from: Quest

a is the answer

explanation:

which statement about trans-fatty acids is false?

a. the conversion of cis-fatty acids to trans-fatty acids is inhibited by the presence of antioxidants.

hope this

Answer from: Quest

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1

0.44 / 0.09 = 4.8

mgco3-5h2o"

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