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>
Boiling point
i hope this helps.
Answer:
C= 0.532M
Explanation:
The equation of reaction is
H2SO4 + 2KOH = K2SO4+ H2O
nA= 1, nB= 2, CA= ?, VA= 48.9ml, CB= 1.5M, VB= 34.7ml
Applying
CAVA/CBVB = nA/nB
(CA× 48.9)/(1.5×34.7)= 1/2
Simplify
CA= 0.532M
Answer:
Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate. Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur.
Explanation:
:)
The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS
