Respuesta :
Based on the mole ratio of the balanced equation of the reactions;
- Gretchen did the experiment incorrectly
- Gretchen did not measure the accurate mass of aluminum required to produce 19 g of aluminum fluoride
- Gretchen needs to react 6.1 g of Al in excess fluorine in order to produce 19 g of AlF₃
What mass of iron(III) oxide is produced from reacting 41 g of Fe with excess oxygen?
The equation of the reaction is given below:
- 4 Fe + 3 O₂ → 2 Fe₂O₃
Based on the reaction, mole ratio of Fe and Fe₂O₃ is 2 : 1
Moles of Fe reacted = 41/56 moles
Mass of Fe₂O₃ produced = 41/56 * 1/2 * 160 = 58.6 g
What mass of aluminum fluoride is produced from reacting 5 g of aluminum with excess fluorine?
The equation of the reaction is given below:
- 2 Al + 3 F₂ → 2 AlF₃
Based on the reaction, mole ratio of Al and AlF₃ is 1 : 1
Moles of Al reacted = 5/27 moles
Mass of AlF₃ produced = 5/27 * 84 = 15.5 g
This is less than the required mass of AlF₃
Moles of AlF₃ in 19 g = 19/84 = 0.2262 moles
Mass of Al required = 0.2262 * 27 = 6.1 g
Based on the results above;
- Gretchen did the experiment incorrectly
- Gretchen did not measure the accurate mass of aluminum required to produce 19 g of aluminum fluoride
- Gretchen need to react 6.1 g of Al in excess fluorine in order to produce 19 g of AlF₃
In conclusion, the mole ratio of the reactions obtained from the equation are required to know the accurate masses of reactants required.
Learn more about mass and mole ratio at: https://brainly.com/question/16806688
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