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ivolga24 [154]
3 years ago
7

If element A has an ionic charge of +2 and element B has an ionic charge of -3, what will be the formula for the compound create

d by element A and element B?
A2B3

A3B2

AB2

impossible to determine without additional information
Chemistry
1 answer:
Softa [21]3 years ago
8 0
A3B2 because the oxidation numbers are the same as ionic charge just switch symbol and number. Then use the cross cross method and you get A3B2.
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An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas
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Answer:

  • The limiting reactant is lead(II) nitrate.
  • 7.20 g of precipitate are formed.
  • 1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

  • Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:

  • 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
  • 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:

  • 0.0296 mol Pb(NO₃)₂ \frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}} * \frac{278.1g}{1molPbCl_{2}} * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

  • 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:

  • 0.0259 mol Pb(NO₃)₂ \frac{2molKCl}{1molPb(NO_{3})_2} * \frac{74.55g}{1molKCl} = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

  • 5.76 - 3.86 = 1.9 g KCl

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