Write the formula for chromium III acetate
1 answer:
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Answer: 60.7 g of will be formed.
Explanation:
To calculate the moles :
The balanced chemical reaction is
is the limiting reagent as it limits the formation of product and
is the excess reagent.
According to stoichiometry :
6 moles of produce = 4 moles of
Thus 2.68 moles of will produce=
of
Mass of
Thus 60.7 g of will be formed by reactiong 60 L of hydrogen gas with an excess of
Answer:
170 g
Explanation:
Moles of Oxygen gas :
Given, Mass of Oxygen = 80.0 g
Molar mass of Oxygen gas= 31.998 g/mol
The formula for the calculation of moles is shown below:
Thus,
The reaction between Al and O₂ is shown below as:
4Al + 3O₂ ⇒ 2Al₂O₃
From the reaction,
3 moles of O₂ on reaction forms 2 moles of Al₂O₃
1 mole of O₂ on reaction forms 2/3 moles of Al₂O₃
2.5 moles of O₂ on reaction forms (2/3)*2.5 moles of Al₂O₃
Moles of Al₂O₃ = 1.6667 moles
Molar mass of Al₂O₃ = 102 g/mol
Mass = Moles * Molar mass = 1.6667 * 102 g = 170 g