What is the mass of aluminum metal that reacts to give 11.1g of manganese metal?
To answer this question, we need to use the balanced chemical equation for the reaction between aluminum and manganese:
2 Al + 3 MnO2 → 3 Mn + Al2O3
The equation shows that 2 moles of aluminum react with 3 moles of manganese dioxide to produce 3 moles of manganese and 1 mole of aluminum oxide.
We can use the molar mass of manganese (54.94 g/mol) and the given mass of manganese (11.1 g) to calculate the number of moles of manganese produced:
moles of Mn = mass of Mn / molar mass of Mn
moles of Mn = 11.1 g / 54.94 g/mol
moles of Mn = 0.202 moles
From the balanced chemical equation, we know that 2 moles of aluminum react with 3 moles of manganese, so we can use the mole ratio to calculate the number of moles of aluminum that reacted:
moles of Al = (2/3) x moles of Mn
moles of Al = (2/3) x 0.202 moles
moles of Al = 0.135 moles
Finally, we can use the molar mass of aluminum (26.98 g/mol) to calculate the mass of aluminum that reacted:
mass of Al = moles of Al x molar mass of Al
mass of Al = 0.135 moles x 26.98 g/mol
mass of Al = 3.65 g
Therefore, the mass of aluminum metal that reacted to give 11.1 g of manganese metal is approximately 3.65 grams.
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