# A sufficient amount of water is added to 49.92 g of NaOH to make 0.600 L of solution. 41.28 g of Al is then added to this solution and hydrogen gas is formed according to the following equation:...

1. A sufficient amount of water is added to 49.92 g of NaOH to make 0.600 L of solution. 41.28 g of Al is then added to this solution and hydrogen gas is formed according to the following equation:

2Al(s) + 6H2O(l) + 2OH-(aq) à 2Al(OH)4- + 3H2(g)

1. How many moles of hydrogen were formed? (Hint: You will need to figure out which one is the limiting reagent)

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First, we need to get the moles and molarity of each of the reactants:

`49.92 grams NaOH * (1 mol e NaOH)/(39.9971 grams NaOH) = 1.248 mol es NaOH = mol es OH^-`

`41.28 grams Al * (1 mol e Al)/(26.981 grams Al) = 1.530 mol es Al`

Next we need to determine the limiting reactant so we can be able to know how many moles of `H_2` gas are produced.

`2Al(s) + 6H_2O(l) + 2OH^(-)(aq) -> 2Al(OH)^(4-) + 3H_2(g)`

`1.248 mol es OH^(-) * (3 mol es H_2)/(2 mol esOH^(-))`

= 1.872 moles `H_2`

`1.530 mol es Al * (3 mol es H_2)/(2 mol es Al)`

= 2.295 moles `H_2`

Looking at the results, we can see that the limiting reactant is the NaOH solution (OH-) and therefore the amount of molels that we will consider is the moles derived from the OH- solution.