Question

Calculate the number of liters of hydrogen, measured at STP, that can be produced from the reaction of 0.155 mol of aluminum according to the following unbalanced chemical reaction equation.


i got .193 L of H2 but it was wrong. i only have one submition left!

Answer 1

You didn't include the unbalanced equation. Is it something like
2Al + 6HCl ==> 3H2 + 2AlCl3 or
2Al + 3H2SO4 ==> 3H2 + Al2(SO4)3
Actually, I think any reaction chosen will be a 3/2 ratio as see below.

0.155 moles Al x (3 moles H2/2 moles Al) = 0.155 x (3/2) = 0.2325 moles H2.
Since one mole occupies 22.4 L at STP so
0.2325mol x 22.4 L/mol = ??

Answer 2

2Al(s) + 2NaOH(aq) + 2H2O(l) 2NaAlO2(aq) + 3H2(g) (unbalanced)

sorry i forgot. it was unbalanced but then i balance it

Answer 3

So my answer didn't change because the ratio still is 3/2.
0.155 moles Al x (3 moles H2/2 moles Al) x (22.4L H2/mol H2) = ??L H2.
I tried to duplicate your error by substituting errors student often make; however, I never could come up with 1.93 L.

Answer 4

thanks so its

5.21 L h2?

Answer 5

thanks for the help: can you help me with another one?

Some sulfuric acid is spilled on a lab bench. It can be neutralized by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows.

2 NaHCO3(s) + H2SO4(aq) --> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)

Sodium bicarbonate is added until the fizzing due to the formation of CO2(g) stops. If 34 mL of 6.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?


i got 5.7 and it said my answer was off by more than 10%

Answer 6

How much H2SO4 was spilled(how many moles?) M x L = 6.0 x 0.035 = 0.204
Convert moles H2SO4 to moles NaHCO3. That will be 0.204 x 2 = 0.408.
Now convert moles NaHCO3 to grams. g = moles x molar mass.