Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
Answer:
D. Nuclei with small masses combine to form nuclei with larger masses.
B. A small amount of mass in the nuclei that combine is converted to energy
Explanation:
A nuclear fusion, in contrary to fission, is the process by which the nuclei of two atoms combine to form a much larger atom with a large nuclei. Likewise, during a fusion reaction, a large amount of energy is released from the small amount of mass in the nuclei (two) that combines.
According to this question, the following are true of a fusion reaction:
- Nuclei with small masses combine to form nuclei with larger masses.
- A small amount of mass in the nuclei that combine is converted to enormous energy.
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
1) The forward reaction is N2 (g) + O2 (g) → 2NO
(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.
2) The equiblibrium equation is
N2 (g) + O2 (g) ⇄ 2NO
3) Then, the reverse reaction is
2NO → N2(g) + O2(g)
Answer: 2NO → N2(g) + O2(g)
