Balanced Eqn
2
C
2
H
6
+
7
O
2
=
4
C
O
2
+
6
H
2
O
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume
60
⋅
300
224
=
80.36
g
ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =
4
⋅
44
⋅
80.36
60
=
235.72
g
and its no. of moles will be
235.72
44
=5.36 where 44 is the molar mass of Carbon dioxide
hope this helps
<span>The two techniques for separating an insoluble solid from a liquid are filtration and centrifuging. Filtration relies on the fact that the solid particles are smaller than the filter paper pores which allow the tiny molecules of liquid to pass through. With solids which dissolve in a liquid solvent, evaporation is commonly used to recover the solid and distillation is used to recover the solvent.</span>
The grams of aluminium extracted from 5000g of alumina is 2647 grams
<h3>Chemical formula of alumina:</h3>
Let's calculate the molecular mass of Al₂O₃
Al₂O₃ = 27 × 2 + 16 × 3 = 54 + 48 = 102 g/mol
Therefore,
102 g of Al₂O₃ = 54 g of aluminium
5000g of Al₂O₃ = ?
mass of aluminium produced = 5000 × 54 / 102
mass of aluminium produced = 270000 / 102
mass of aluminium produced = 2647.05882353
mass of aluminium produced = 2647 grams
learn more on mass here: brainly.com/question/14627327
Answer:
So first thing to do in these types of problems is write out your chemical reaction and balance it:
Mg + O2 --> MgO
Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.
To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.
The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.
The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.
Percent yield is acutal/theoretical, .66/.693, or 95.24%.
I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.
Hope this helps.
There are a few things you can do to make slime less sticky. You can either:
1) add a bit of baking soda (one to three tea spoons depending on how big it is, for the one that you show I would say one should be enough)
2)You can place it in hot water and squish the water out of it before placing it in cold water and then drying it out a bit with a towel. Don't forget to knead it a bit so it stays smooth and not too hard.
