equation 2Na + 2H 2 O -------> 2NaOH + H 2 if 90.0 grams of sodium is dropped in
ID: 675638 • Letter: E
Question
equation 2Na + 2H2O -------> 2NaOH + H2 if 90.0 grams of sodium is dropped into 80.0 ml of water, howmany liters of hydrogen at STP woud be produced? which reactant is in excess and how much of it is leftover? equation 2Na + 2H2O -------> 2NaOH + H2 if 90.0 grams of sodium is dropped into 80.0 ml of water, howmany liters of hydrogen at STP woud be produced? which reactant is in excess and how much of it is leftover? if 90.0 grams of sodium is dropped into 80.0 ml of water, howmany liters of hydrogen at STP woud be produced? which reactant is in excess and how much of it is leftover?Explanation / Answer
Hi I hope this helps. 1. First List all of the givens, and balance thechemical equation if is not already done so. 90g Na 80 mL H2O. Standard Temperature and Pressure (STP) refers tothe ideal gas law PV=nRT under standard conditions means Temperature (T)= 298 K and Pressure (P)= 1 atm n=numberof moles R= .0821 L atm/K mol 2. Think about what you are looking for and how you aregetting there. Final answer is : ? Liters of H2 gas. 3. The balanced reaction enables you to get from onecomponent to another. BUT when you have TWO givens then it becomes a LimitingReactant problem. With any Chemistry problem you cannot workwith anything unless it is in moles, so the first thing is tochange to moles. 90g Na x 1 mole Na = 3.91 moles Na 23 gNa 80 mL H2O x 1gH2O = 80 gH2O 1mLH2O 80g H2O x 1mol H2O =4.44 moles H2O 18g H2O 4. Now that you have determined moles of each given youneed to figure out the Limiting reagent. (a) Do this by dividing by the smallest number of moles to geta ratio between the two. 3.91 mol Na /3.91 = 1/ 2 = .5 Limiting Reactant (smallest) 4.44 mol H2O / 3.91 = 1.14/2 = .57 Excess(largest) (b) Then divide by the corresponding coefficient from thebalanced reaction and the smallest number is the LimitingReactant. 5. Now you know which reactant to work with (Na) so nowyou are ready to find the unknown. Be very careful to use theactual number of moles of Na and not all the other math wedid. 3.91 mol Na -------> mol H2 6. In order to convert from moles of Na to mol H2you need to use the mole to mole ratio from the balanced reaction(corresponding coefficients) 3.91 mol Na x 1 mol H2 = 1.955mol H2 2 mol Na 7. We would be done if the answer was in moles but they wantLiters so we have to plug in the number of moles of hydrogen intothe ideal gas law. P= 1 atm, V= ?, n= 1.955 mol , R= .0821 Latm/ mol K , T= 298 K 2 mol Na 7. We would be done if the answer was in moles but they wantLiters so we have to plug in the number of moles of hydrogen intothe ideal gas law. P= 1 atm, V= ?, n= 1.955 mol , R= .0821 Latm/ mol K , T= 298 K You may notice that although the final anwer is in Liters ofHydrogen we had to first find the number of moles, so this was atwo part problem.Related Questions
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