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Darina [25.2K]
3 years ago
12

When the following equation is balanced using the smallest possible integers, what is the number in front of the substance in bo

ld type?
Al + Fe3O4→ Al2O3 + Fe
Chemistry
1 answer:
weeeeeb [17]3 years ago
7 0

The balanced equation :

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

You just have to look for which element is in bold because the question is not clear

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

Al + Fe₃O₄→ Al₂O₃ + Fe

give a coefficient

aAl + Fe₃O₄→ bAl₂O₃ + cFe

O, left=4, right=3b⇒3b=4⇒b=4/3

Al, left=a, right=2b⇒a=2b⇒a=2.4/3⇒a=8/3

Fe, left=3, right=c⇒c=3

The equation becomes :

8/3Al + Fe₃O₄→ 4/3Al₂O₃ + 3Fe x3

8Al + 3Fe₃O₄→ 4Al₂O₃ + 9Fe

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Natalka [10]

<u>Answer:</u> The molarity of Cr^{3+} ions in the solution is 0.299 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For chromium (III) acetate:</u>

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol

1 mole of chromium (III) acetate (Cr(CH_3COO)_3) produces 1 mole of chromium (Cr^{3+}) ions and 3 moles of acetate (CH_3COO^-) ions

Moles of Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles

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Molarity of chromium (III) nitrate solution = 0.461 M

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Putting values in equation 1, we get:

0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol

1 mole of chromium (III) nitrate (Cr(NO_3)_3) produces 1 mole of chromium (Cr^{3+}) ions and 3 moles of nitrate (NO_3^-) ions

Moles of Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles

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Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

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\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+} ions in the solution is 0.299 M

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