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How many moles are present in 32.5 g aluminum chloride?

Answer 1

Naomi Aronson

#"Moles of aluminum chloride "=" approx. 0.250 moles"#

#"Moles"# #=# #"Mass of compound"/"Molar mass of compound"#

#=# #(32.5*g)/((26.98+3xx35.45)*g*mol^-1)#

#=# #(32.5*g)/(133.33*g*mol^-1)# #=# #??" moles"#.

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Answer 2

Olivia Anton

The formula to determine how many moles are in 32.5 g of aluminum chloride is moles = mass / molar mass. The molar mass of aluminum chloride (AlCl3) is the product of the atomic masses of aluminum (Al) and three times the atomic mass of chlorine (Cl). The atomic mass of aluminum (Al) is approximately 26.98 g/mol, while the atomic mass of chlorine (Cl) is approximately 35.45 g/mol. As a result, the molar mass of aluminum chloride (AlCl3) is 26.98 g/mol (Al) + 3 * 35.45 g/mol (Cl) = 133.33 g/mol. From there, we can determine how many moles are in 32.5 g of aluminum chloride: moles = 32.5 g / 133.33 g/mol ≈ 0.244 moles.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.