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djverab [1.8K]
3 years ago
11

Which of the following compounds contains the Lead (II) ion? PbO PbCl4 Pb2O Pb2S

Chemistry
2 answers:
KATRIN_1 [288]3 years ago
6 0
<span>PbO Let's look at each of the 4 compounds and see what's needed. PbO. * Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice. PbCl4 * Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice. Pb2O * Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice. Pb2S * Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>
nikitadnepr [17]3 years ago
4 0
1) A
2) B
3) A ---->your question
4) C
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6.0 mol Al reacts with 4.0 mol O2 to form Al2O3.
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3.0 moles Al₂O₃

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We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.

4 Al + 3 O₂ -----> 2 Al₂O₃

6.0 moles Al            2 moles Al₂O₃
----------------------  x  -------------------------  =  3.0 moles Al₂O₃
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4.0 moles O₂           2 moles Al₂O₃
----------------------  x  -------------------------  =  2.7 moles Al₂O₃
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As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.

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