Gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 18. g of ethane is mixed with 85.4 g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. round your answer to 2 significant digits.

The complete balance chemical reaction for the compounds stated in the problem is:

C2H6 + (7/2) O2 ---> 2 CO2 + 3 H2O

First, let us find out which is the limiting reactant. We can know by calculating for the moles of reactant.

moles C2H6 = (18 g) (1 mol / 30.07 g) = 0.5986 mol

moles O2 = (85.4 g) (1 mol / 32 g) = 2.6688 mol

The ratio in the chemical reaction is 3.5O2:1C2H6, the given chemicals are in the ratio of:

2.6688O2:0.5986C2H6 = 4.46O2:1C2H6

Therefore this means that there is excess O2 and the C2H6 is our limiting reactant. Therefore we calculate the amount of CO2 based on this.

mass CO2 = (0.5986 mol C2H6) (2 mol CO2/ 1 mol C2H6) (44 g CO2/ mol)

mass CO2 = 52.677g

In 2 significant digits:

mass CO2 = 53 g

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explanation:

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).