Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
Answer:
1.Very good electrical conductivity :<u> Metals</u> (Decreacing order of conductivity)
- <em>Silver > Copper > Gold > aluminium</em>
2. Amphoteric <u>: Metal elements</u>
- <em>Beryllium , Aluminium , Zinc </em>,
3.Gaseous at room temperature: mostly <u>Nobel gases elements</u> and some non - metal elements.
- <em>Helium ,neon , argon , krypton , fluorine , Oxygen , nitrogen</em>
4.Solid at room temperature:<u> Mostly Metals</u> (few non-metals, metalloid elements)
- <em>Metals (Sodium , potassium , calcium , gold are solid)</em>
<em>Non- metals(Carbon ,Boron )</em>
<em>Metalloids(antimony)</em>
<em>5.</em> Brittle <em>: </em><u>non - metals </u>(can't be rolled into wires)
<em>Hydrogen , carbon , sulfur , phosphorus</em><u> </u>
Explanation:
Answer:
true!
Explanation:
if youre going by the scientific method, observations are mainly first. then questions, research, etc
calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)
