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3 years ago

Fe2O3 + 2A1 - Al2O3 + 2Fe

1 answer:

4 0

I’m assuming in this scenario that 2.5 mols of Fe2O3 is reacting with excess reactant. Based on the balanced equation, you can see that there’s a 1:2 ratio between Fe2O3 and Fe. This means that to find the mols of Fe formed, you need to multiply the mols of Fe2O3 by 2.
2.50mol Fe2O3 * 2mol Fe/1 mol of Fe2O3 = 5.00mols Fe.

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3 years ago

How are atoms considered the “alphabet” in language of molecules?

luda_lava [24]

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3 years ago

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a (n) bec

galben [10]

Answer:

C) cation, loses, anion, gains

Explanation:

Lithium is a metal from Group 1, so it has 1 valence electron. Thus, it loses 1 electron to complete its octet and form the cation Li⁺.

Oxygen is a nonmetal from Group 16, so it has 6 valence electrons. Thus, it gains 2 electrons to complete its octet and form the anion O²⁻.

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a cation because it loses an electron, and oxygen would be an anion because it gains 2 electrons.

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4 years ago

Why is the atom neutrally charged?

V125BC [204]

Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged).

3 0

3 years ago

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

Vladimir [108]

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

Second, we gather the information what we are going to use in our calculations.

Volumen of Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 = 0.541M

Molecular Weight Al2O3 = 101.96 g/mol

Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

$\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 = 0.022moles Al(NO3)3$

then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced

$0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3$

finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.

$0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3$

4 0

4 years ago