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Luden [163]
4 years ago
9

30 pts! Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liter

s of hydrogen gas are formed at STP? Show all work for credit. 2Na+2H2O--> 2NaOH+H2
Chemistry
1 answer:
mihalych1998 [28]4 years ago
3 0
First, calculate the number of moles of sodium present with the given mass,

              31.5 g of sodium x (1 mol sodium/ 23 g sodium) = 1.37 mol sodium

It is given in the equation that for every 2mols of sodium, one mol of H2 is produced.

           mols of H2 = (1.37 mols sodium)(1 mol H2/ 2 mols sodium)
                  mols of H2 = 0.685 mols H2

Then, at STP, 1 mol of gas = 22.4 L.
                
                 volume of H2 = (0.685 mols H2)(22.4 L / 1 mol)
                     volume of H2 = 15.34 L

Answer: 15.34 L
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Solution : Given,

For Accelerator 1 model,

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For Accelerator 2 model,

Input energy = 7690.0 J

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Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

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Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy )  ×  100

For Accelerator 1 model,

% energy lost as waste = \frac{663.1}{2078.3}\times100 = 31.90%

For Accelerator 2 model,

% energy lost as waste = \frac{2337.5}{7690.0}\times100 = 30.39%

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