Count the atoms 2Ca(H2PO4)3 How many O atoms are in the above equation?
1 answer:
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Answer:
4.18 g
Explanation:
The formula for the calculation of moles is shown below:
Given: For Li
Given mass = 2.50 g
Molar mass of Li = 6.94 g/mol
<u>Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles</u>
Given: For
Given mass = 2.50 g
Molar mass of = 28.02 g/mol
<u>Moles of = 2.50 g / 28.02 g/mol = 0.08924 moles</u>
According to the given reaction:
6 moles of Li react with 1 mole of
1 mole of Li react with 1/6 mole of
0.3602 mole of Li react with mole of
Moles of that will react = 0.06 moles
Available moles of = 0.08924 moles
is in large excess. (0.08924 > 0.06)
Limiting reagent is the one which is present in small amount. Thus,
Li is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
6 moles of Li gives 2 mole of
1 mole of Li gives 2/6 mole of
0.3602 mole of Li react with mole of
Moles of = 0.12
Molar mass of = 34.83 g/mol
Mass of = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g
<u>Theoretical yield = 4.18 g</u>