gaseous ethane (ch, ch) will react with gaseous oxygen (o2) to produce gaseous carbon dioxide (co2) and gaseous water (h2o). suppose 24 g of
ethane is misced with 16.9 g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to 2
significant digits.

Mass =  8.1 g

Explanation:

Given data:

Mass of ethane = 24 g

Mass of oxygen = 16.9 g

Mass of water = ?

Solution:

Chemical equation:

2C₂H₆ + 7O₂  → 4CO₂ + 6H₂O

Moles of ethane:

Number of moles = mass/molar mass

Number of moles = 24 g / 30.07 g/mol

Number of moles = 0.8 mol

Moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 16.9 g / 32 g/mol

Number of moles = 0.53 mol

Now we will compare the moles of water with oxygen and ethane.

                         C₂H₆            :             H₂O

                            2               :                6

                         0.8               :           6/2×0.8 = 2.4 mol

                          O₂               :             H₂O

                            7                :              6

                            0.53           :             6/7×0.53= 0.45 mol

It means oxygen is limiting reactant because moles of water produced by given amount of oxygen are less.

Mass of water:

Mass =  number of moles × molar mass

Mass =  0.45 mol ×  18 g/mol

Mass =  8.1 g

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g