Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
6.02 x 1023 atoms weigh out 63.55 grams copper.
No. of Molecules in water = 3.5mole x (6.02 x 10^23) molecules/mole = 2.107 x 10^24 molecules of H2O
Answer:
Explanation:
given volume =5.3 litres
as we know that 1 litre =1 dm3
therefore 5.3 litres =5.3 dm3
now moles =given mass /molar mass
so mass =moles*molar mass
in order to find mass of carbon we need to find moles
therefore moles=given volume/standard volume standard volume at STP is 22.4 dm3
moles= 5.3/22.4
moles=0.236=0.24
now mass of carbon dioxide =moles*molar mass
mass = 0.24*44
mass=10.56kg
4Fe(s) + 3O2(g) → 2Fe2O3(s) ....... ΔH = -1.7 x 10^3 kJ
<span>shows that 1700 kJ of heat is released when 4 moles of Fe react. </span>
<span>Molar mass of Fe = 56 g/mol </span>
<span>Molar mass of Fe2O3 = 160 g/mol. </span>
<span>(a) When 4 mol ( 4 x 56 = 224 g) iron reacted 1700 kJ are released, </span>
<span>when 9.0 g Fe reacted, the amount of energy will be </span>
<span>(9.0 x 1700) / (224) = 68.3 kJ </span>
<span>(b) According to the equation, 1700 kJ of energy released during the formation of 2 moles of Fe2O3 which is 2 x 160 = 320 g of rust. </span>
<span>120 kcal x (4.18 kJ / 1 kcal) = 501.6 kJ </span>
<span>When 1700 kJ released = 320 g rust </span>
<span>when 501.6 kJ released = (501.6 x 320) / 1700 = 94.4 g rust</span>
Reduction takes place in cathode while oxidation takes place at the anode.
Given that the reaction 3MnO4- +24H + +5Fe→3Mn+2+5Fe+3+12H2O
Now that oxidation is termed as an increase in oxidation number or loss of electrons while reduction is a decrease in oxidation number and gain of electrons,
∴oxidation will be
Fe→Fe +3+3e-
reduction will
MnO4- +8H+ →Mn+2+4H2O
If oxidation takes place at anode then Anode: an oxidation reaction
Fe→Fe+3+3e- . Then the answer is
Fe(s)→Fe+3(aq)+3e-