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You need the set of reactions that goes from ammonia to nitric acid.
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1) 4NH3(g)+5O2(g)-->4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)-->2NO2(g)
3) 3NO2(g)+H2O(l)-->2HNO3(aq)+NO(g)
State the ratio of moles of HNO3 to NH3:
4 moles of NH3 produce 4 mole of NO,
4 moles of NO produce 4 moles of NO2
4 moles of NO2 produce 4 * (2 / 3) moles of HNO3 = 8/3 moles of HNO3.
=> (8/3) moles HNO3 : 4 moles NH3
Calculate the number of moles of HNO3 in 900.00 l of 0.140 M solution
M = n / V => n = M * V = 0.140 M * 900.00 liter = 126 moles HNO3
Use proportions:
(</span><span>8/3) moles HNO3 / 4 moles NH3 = 126 moles HNO3 / x
=> x = 126 moles HNO3 * 4 moles NH3 / (8/3 moles HNO3) = 189 moles NH3
Convert moles to grams:
molar mass NH3 = 14 g/mol + 3 * 1g/mol = 17 g/mol
mass in grams = number of moles * molar mass = 189 moles * 17 g/mol = 3213 g
Answer: 3213 g.
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Answer:
Pressure for H₂ = 11.9 atm
Option 5.
Explanation:
We determine the complete reaction:
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
As we do not know anything about the HCl, we assume that the limiting reactant is the Al and the acid is the excess reagent.
Ratio is 2:3.
2 moles of Al, can produce 3 moles of hydrogen
Therefore 4.5 moles of Al must produce (4.5 . 3) / 2 = 6.75 moles
Now we can apply the Ideal Gases law to find the H₂'s pressure
P . V = n . R . T → P = (n . R .T) / V
We replace data: (6.75 mol . 0.082L.atm/mol.K . 300K) / 14L
Pressure for H₂ = 11.9 atm
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Explanation:
