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3 years ago

4Fe + 3O2 → 2Fe2O3

Chemistry

1 answer:

topjm [15]3 years ago

8 0

Answer:

moles Fe₂O₃ = 0.938 mole

Explanation:

Convert given data to moles, solve in terms of moles by reaction ratios in balanced equation. After obtaining answer in moles, convert to needed dimension. In this case, no conversions are needed.

4Fe + 3O₂ => 2Fe₂O₃

105g/56g·mol⁻¹

= 1.875 mol Fe => => => => => => 2/4(1.875 mol Fe₂O₃) = 0.938 mol Fe₂O₃

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$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

$Mg$ ⇒ $Mg^{2+}+2e^-$

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Explanation:

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

3 years ago

Read 2 more answers

Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.

STALIN [3.7K]

Answer: Solution A : $[H_3O^+]=0.300\times 10^{-7}M$

Solution B : $[OH^-]=0.107\times 10^{-5}M$

Solution C : $[OH^-]=0.177\times 10^{-10}M$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log[H_3O^+]$