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A sufficient amount of water is added to 49.92 g of NaOH to make 0.600 L of solution....
- 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)
- How many moles of hydrogen were formed? (Hint: You will need to figure out which one is the limiting reagent)
1 Answer | add yours
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.
Final answer: 1.872 moles H2
Posted by jerichorayel on April 23, 2013 at 2:27 AM (Answer #1)
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