How many oxygen atoms in 4 Al 2 O 3
1 answer:
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Answer:
Explanation:
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In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
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Answer:
Explanation:
First write the balance eqation of chemical reaction:
Remember writing any chemical reaction from its elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.
mass of oxygen given=5gram
mole of oxygen
mass of sulpher given=6gram
mole of sulpher
from the above balanced equaion;
2 mole of sulphur reacts with 3 mole Oxygen completely
1 mole of sulphur reacts with 1.5 mole Oxygen completely
0.19 mole of sulphur reacts with i.e. 0.285 mole Oxygen completely.
but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent
so product will depend on the limiting reagent
from the balance equation
3 mole of sulpher gives 2 mole
1 mole of sulpher will give 2/3 mole
0.16 of sulpher will give 0.12 mole
mass of =9.6gram this is theoreical production of
and
actual production of =7.9gram
Explanation:
Molarity is defined as number of moles per liter of solution.
Mathematically, molarity =
It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.
molarity =
0.0800 M =
no. of moles = 1.6 mol
Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.
No. of moles =
mass in grams =
=
= 399.488 g
Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.