The chemical reaction for this is:
2 C2H6 + 7 O2 => 4 CO2 + 6 H2O
Solving for CO2 with each reactant will give:
21.0 g C2H6 x (1 mol C2H6/30.08 g C2H6) x (6 mol H2O/2
mol C2H6) x (18 g H2O/1 mol H2O) = 37.70 g H2O
110 g O2 x (1 mol O2/32.00 g O2) x (6 mol CO2/7 mol O2) x
(18 g H2O/1 mol H2O) = 53.04 g H2O
Since the amount of H2O in C2H6 is lower therefore C2H6
is the limiting reactant and the maximum amount of water is only 38 g H2O (2 significant digits)
ANswer:
38 g water
Answer:
41.16 moles of H2O
Explanation:
Ratio for the products-reactants is 1:6, so 1 mol of glucose is produced when plants use 6 moles of water.
Then, let's make a rule of three:
1 mol of glucose is produce by using 6 moles of water
6.86 moles of glucose are produced by the use of (6 . 6.86)/1 = 41.16 moles of H2O
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
