Answer:
Theoretical yield of vanadium = 1.6 moles
Explanation:
Moles of 
= 1.0 moles
Moles of 
= 4.0 moles
According to the given reaction:-
1 mole of react with 5 moles of
Moles of Ca available = 4.0 moles
Limiting reagent is the one which is present in small amount. Thus, Ca is limiting reagent. (4.0 < 5)
The formation of the product is governed by the limiting reagent. So,
5 moles of Ca on reaction forms 2 moles of V
1 mole of Ca on reaction for 2/5 mole of V
4.0 mole of Ca on reaction for 
mole of V
Moles of V = 1.6 moles
<u>Theoretical yield of vanadium = 1.6 moles</u>
In order to balance this, you have to count
each element where the elements in the reactants side and the product side
should have equal number of molecules. The balanced reaction is as follows:
KOH + H3PO4 = KH2PO4 +H2O
Answer:
Lipids and protein.
Explanation:
The cell membrane is mostly made up of lipids and protein. It is made up of other things, but it is mostly made up of those.
I hope it helps! Have a great day!
bren~
Answer:
1461.7 g of AgI
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:
From the balanced equation above,
1 mole of CaI₂ reacted to produce 2 moles of AgI.
Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI
Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:
Mole of AgI = 6.22 moles
Molar mass of AgI = 108 + 127
= 235 g/mol
Mass of AgI =?
Mass = mole × molar mass
Mass of AgI = 6.22 × 235
Mass of AgI = 1461.7 g
Therefore, 1461.7 g of AgI were obtained from the reaction.
