You work it the same way you do any stoichiometry problem.
You know Na is the limiting reagent, it will be completely consumed; therefore, how much H2O must be used according to the equation? It must be
4.00moles Na x (2 moles H2O/2 moles Na) = 4.00 x (1) = 4.00 moles H2O. You had 4.2; therefore, 0.2 mols H2O must remain un-reacted. How many grams is that?
0.2 x 18 = 3.6 grams.
76.0-3.60 = ?? that didn't react.(amount not consumed).
Now that you see how to do that you probably can do the second problem. moles HNO3 - moles KOH = moles in excess.
Please help me with these questions:
1. Consider the equation:2Na+ 2H2O-->2NaOH +H2
-If 92.0g of sodium is reacted with 76.0g of water until the reaction goes to completion, which reactant will remain and in what quantity?
A: I know that H20 is the reactant that is in excess with 4.2 moles whereas Na has 4.0 moles.-But how do you find the quantity?
2. 12.5 ml of 0.280M HNO3 and 5.0mL of 0.920M KOH are mixed. To make the resulting solution neutral, which of the following should be added?
-1.1 mol HCl
-1.69 mL of 0.650M HBr
-0.55 mmol Mg(OH)2
-2.2mL of 0.50M LiOH
-none of the above; already neutral
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