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stealth61 [152]
2 years ago
13

If 5.10 g of sodium and 305 g of potassium nitrate react in an airbag, how many grams of kno3 remain because of the limited amou

nt of sodium present? 10na(s) 2kno3(s) k2o(s) 5na2o(s) n2(g)
Chemistry
1 answer:
Oksi-84 [34.3K]2 years ago
7 0
The balanced equation for the above reaction is as follows;
<span>10Na(s) +  2KNO</span>₃(s)-->  K₂O(s) +  5Na₂O(s) +  N₂<span>(g)
Stoichiometry of Na to KNO</span>₃ is 10 : 2
Number of Na moles reacted - 5.1 g/ 23 g/mol = 0.22 mol
Na is the limiting reactant, therefore Na is fully used up, Since KNO₃ is present in excess , at the end of the reaction a certain amount of KNO₃ will be remaining.
10 mol of Na reacts with 2 mol of KNO₃
Therefore 0.22 mol of Na reacts with - 2 /10 x 0.22 = 0.044 mol of KNO₃
Mass of KNO₃ reacted = 0.044 mol x 101.1 g/mol = 4.45 g
Mass of KNO₃ present initially - 305 g
Therefore remaining mass of KNO₃  - 305 - 4.45 = 300.55 g

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Initial                               0.1 M     ⇆    0          0
At Equilibrium                 0.1-X      ⇆    X         X
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(www.cymath.com)
                                 X  =  0.028 M

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                                 =  28 %     (Percentage Ionization)
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