How many moles of calcium oxide will be produced when 1.6 moles of Iron(III)oxide reacts with calcium phosphate according to the
1 answer:
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Answer:
72.03 %
Explanation:
Total mass of dolomite = 9.66 g
Let the mass of Magnesium carbonate = x g
The mass of calcium carbonate = 9.66 - x g
Calculation of the moles of Magnesium carbonate as:-
Molar mass of Magnesium carbonate = 122.44 g/mol
The formula for the calculation of moles is shown below:
Thus,
Calculation of the moles of calcium carbonate as:-
Molar mass of calcium carbonate = 100.0869 g/mol
Thus,
According to the reaction shown below:-
In both the cases, the oxides formed from the carbonates in the 1:1 ratio.
So, Moles of MgO =
Molar mass of MgO = 40.3044 g/mol
Thus, Mass = Moles*Molar mass =
Moles of CaO =
Molar mass of CaO = 56.0774 g/mol
Thus, Mass = Moles*Molar mass =
Given that total mass of the oxide = 4.84 g
Thus,
Thus, the mass of Magnesium carbonate = 6.9582 g
Number of moles : n₂ = 1.775 moles
<h3>Further explanation</h3>Given
Moles = n₁ = 1.4
Volume = V₁=22.4 L
V₂=28.4 L
Required
Moles-n₂
Solution
Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
The ratio of gas volume will be equal to the ratio of gas moles
Input the values :
n₂ = (V₂ x n₁)/V₁
n₂ = (28.4 x 1.4)/22.4
n₂ = 1.775 moles