88g of sodium reacts with 62 grams of oxygen to produce sodium oxide.
a. Which is the limiting reagent?
b. How much product will be formed in grams and moles?
c. If the actual yield is 38 grams, what is the percentage yield?
Limiting reagent. PLEASE HELP!!!!?
2014-07-22 5:42 am
回答 (1)
2014-07-22 8:00 am
4 Na + O2 → 2 Na2O
(88 g Na) / (22.98977 g Na/mol) = 3.828 mol Na
(62 g O2) / (31.99886 g O2/mol) = 1.938 mol O2
a.
3.828 moles of Na would react completely with 3.828 x (1/4) = 0.957 mole of O2, but there is more O2 present than that, so O2 is in excess and Na is the limiting reagent.
b.
(3.828 mol Na) x (2 mol Na2O / 4 mol Na) x (61.97894 g Na2O/mol) = 118.63 = 119 g Na2O
c.
(38 g) / (118.63 g) = 0.32 = 32% yield
(88 g Na) / (22.98977 g Na/mol) = 3.828 mol Na
(62 g O2) / (31.99886 g O2/mol) = 1.938 mol O2
a.
3.828 moles of Na would react completely with 3.828 x (1/4) = 0.957 mole of O2, but there is more O2 present than that, so O2 is in excess and Na is the limiting reagent.
b.
(3.828 mol Na) x (2 mol Na2O / 4 mol Na) x (61.97894 g Na2O/mol) = 118.63 = 119 g Na2O
c.
(38 g) / (118.63 g) = 0.32 = 32% yield
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