The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
Answer;
=28.09 amu
Explanation;
In this problem, they did not give us the percentages. However, since we know the number of atoms, we can easily calculate the percentages. For example:
(460 X 100)/500 = 92%
If we do this for all three isotopes,
(460 × 25)/500 = 5 %
(460 × 15) /500 = 3%
-We get 92%, 5%, and 3%. (We'll assume these are absolute numbers for determining our significant figures).
Now the problem is just like the previous one. First convert the percentages into decimals. Then multiply those decimals by the masses and add. Here's the solution:
= (0.92) X (27.98 amu) + (0.05) X (28.98 amu) + (0.03) X (29.97 amu)
= 25.74 amu + 1.449 amu + 0.8991 amu
= 28.09 amu
I would think the answer would be C. Based off of what I've learned and heard. Sorry if it's not the correct answer though.
