Julie S.
asked • 03/01/20How many grams of table salt (sodium chloride) can be produced from 10 g of sodium and 12.5 grams of chlorine gas?
Label excess and limiting reactants.
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1 Expert Answer
J.R. S. answered • 03/02/20
Tutor
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Ph.D. University Professor with 10+ years Tutoring Experience
2Na(s) + Cl2(g) ==> 2NaCl(s) ... balanced equation
Find limiting reactant:
moles Na = 10 g Na x 1 mol Na/23 g = 0.43 moles Na
moles Cl2 = 12.5 g Cl2 x 1 mol Cl2/71 g = 0.18 moles Cl2
Since the balanced equation shows a 1:1 mole ratio of Na and Cl2, the limiting reactant is Cl2
This means that the excess reactant is NaCl
Find grams of NaCl produced from 0.18 moles of Cl2:
0.18 moles Cl2 x 2 moles NaCl/1 mole Cl2 x 58 g NaCl/mol NaCl = 20.9 g NaCl = 20 g NaCl (to 1 sig. fig.)
Julie S.
J.R., thank you so much for your help! :)
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03/03/20
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Stanton D.
1) Convert g->moles using atomic weight (23) or molecular weight (71) of reactants respectively . 2) Write balanced equation for rxn. 3) Examine data from step (1) to see which reactant is limiting. (you should be able to see that it is Cl2). 4) Take that limiting mol of reactant "through" the reaction to product, using the rxn stiochiometry. 5) convert that product back to g, using FW of NaCl. -- Cheers, -- Mr. d.03/02/20