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photoshop1234 [79]
3 years ago
15

Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu

mber of moles of al, s, and o atoms numerically, separated by commas.
Chemistry
2 answers:
earnstyle [38]3 years ago
8 0
<span>Answer: 8, 12, 48.
</span>

<span>Explanation:
</span>

<span>1) Aluminum:
</span>

<span>There are 2 atoms of Al in the unit formula Al₂(SO₄)₃, so there are 4x2 = 8 atoms of Al in 4 unit formulas of the same compound.
</span>

<span>2) Sulfur:
</span>

There are 3 atoms of S in the unit formula <span>Al₂(SO₄)₃, so there are 4x3 = 12 atoms of S in 4 unit formulas of the same compound
</span>

3) Oxygen:

<span>
</span><span>There are 4x3 = 12 atoms of O in the unit formula </span><span>Al₂(SO₄)₃, so there are 4x12 = 48 atoms of O in 4 unit formulas of the same compound.</span>


So, the answer is 8, 12, 48
bagirrra123 [75]3 years ago
3 0
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.

So the answer is:
8,12,48


</span>
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Molarity = number of moles of solute / L of solution

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7 0
3 years ago
4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

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