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krek1111 [17]
3 years ago
8

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

three nitrate and lead
Chemistry
1 answer:
Pepsi [2]3 years ago
7 0

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as \text{Pb}^{2+} ions. The formula for a nitrate ion is {\text{NO}_3}^{-}. The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each \text{Pb}^{2+} ion in lead nitrate would pair up with two {\text{NO}_3}^{-} ions. The formula for lead nitrate will be \text{Pb}({\text{NO}_3})_2. Each formula unit of lead nitrate will contain one \text{Pb}^{2+} ion and two {\text{NO}_3}^{-} ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as \text{Cr}^{3+}. Similarly, each \text{Cr}^{3+} will pair up with three {\text{NO}_3}^{-} ions. The formula for chromium (III) nitrate will be \text{Cr}(\text{NO}_3})_3. Each formula unit of chromium (III) nitrate will contain one {\text{NO}_3}^{-} ion and three {\text{NO}_3}^{-} ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of {\text{NO}_3}^{-} ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

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Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

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Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

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The moles of K₃PO₄ are:

13,9 g of K₃PO₄ ×\frac{1mol}{212,27 g} = 0,0655 moles of K₃PO₄

The molarity of the solution is:

\frac{0,0655 moles}{0,25L} = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

\frac{0,1965 moles}{0,25L} = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

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