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4 years ago

Did Pb2+ react with KI? Yes or No? State the reason of your answer.

Chemistry

2 answers:

Snowcat [4.5K]4 years ago

8 0

Answer: Yes, lead ion will react with potassium iodide.

Explanation: When lead ion reacts with KI, a bright yellow colored precipitate is formed.

The reaction of these two follows:

$Pb^{2+}(aq.)+2KI(aq.)\rightarrow PbI_2(s)+2K^+(aq.)$

Hence, $Pb^{2+}$ ion reacts with KI to yield $PbI_2$

This test is the confirmatory test for lead ion which is present in Group I.

madam [21]4 years ago

4 0

Yes, Pb2+ react with KI.

Pb2+ is lead (II) ion
KI is a compound. It is a combination of potassium (K) and iodine (I). It is called potassium iodide.

Both potassium and iodine are highly reactive elements. Thus, when they are combined with another element like Pb2+, Pb2+ reacts with KI

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4 years ago

If 175 g of phosphoric acid reacts with 150.0 g of sodium hydroxide, what is the limiting reactant? How many grams of sodium pho

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Answer:

NaOH is the limiting reactant.

204.9 g of sodium phosphate are formed.

51.94 g of excess reactant will remain.

Explanation:

The reaction that takes place is:

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NaOH ⇒ 150 g ÷ 40 g/mol = 3.75 mol

1.78 moles of H₃PO₄ would react completely with (1.78 * 3) 5.34 moles of NaOH. There are not as many NaOH moles so NaOH is the limiting reactant.

We calculate the produced moles of Na₃PO₄ using the limiting reactant:

3.75 mol NaOH * $\frac{1molNa_3PO_4}{3molNaOH}$ = 1.25 mol Na₃PO₄

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1.25 mol Na₃PO₄ * 163.94 g/mol = 204.9 g

We calculate how many H₃PO₄ moles would react with 3.75 NaOH moles:

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3 years ago

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Answer:

The rate law is [B]

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Similarly, if we look at trials 2 and 3, the concentration of B is constant, while A is changing. In this case, the rate has not changed at all with a change in concentration of A, so this means that A has 0 order.

Therefore, the rate law is just [B].

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