4NH3 + 5O2 ==> 4NO + 6H2O Balanced equation
ALWAYS WORK IN MOLES, NOT IN GRAMS
moles of NO produced = 70.5 g NO x 1 mole/30 g = 2.35 moles NO
Since this represents only a 29.8% yield, find what 100% yield would be:
2.35 moles/0.298 = 7.89 moles of NO
From the balanced equation 4 moles NH3 produces 4 moles of NO. Calculate moles of NH3 needed:
7.89 moles NO x 4 moles NH3/4 moles NO = 7.89 moles NH3 needed
Find grams of NH3 needed:
7.89 moles NH3 x 17 g/mole = 134 g NH3 needed
Answer:
Length
- centimetre
- millimetre
- kilometre
- meter
Mass
- gram
- kilogram
Capacity
- liter
- millilitre
I don't know why mass is there but hope this helps! :)
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
They went from atoms to oxygen molecule.
Explanation:
I'm not sure if this is the answer your teacher is looking for, but in simple terms you had 2 oxygen atoms. Together they make an oxygen molecule, which is the stuff we breathe.
