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9.42 g of lead (ii) nitrate was reacted with 0.655 mol/L potassium iodide to produce 2.64 g of precipitate. What volume of potas

sium iodide was used in this reaction?

1 answer:

Artist 52 [7]1 year ago

4 0

The volume of potassium iodide used will be 0.087 L or 87 mL

<h3>Stoichiometric calculation</h3>

The equation of the reaction goes thus: $Pb(NO_3)_2(aq) + 2KI --- > PbI_2(s) + 2KNO_3(aq)$

The mole ratio of the 2 reactants is 1:2.

Mole of 9.42 g $Pb(NO_3)_2$ = 9.42/331.2 = 0.02844 mol

Equivalent mole of KI = 0.02844 x 2 = 0.05688 mol

Volume of 0.05688 mol, 0.655 mol/L KI = 0.05688/0.655 = 0.087 L or 87 mL

More on stoichiometric problems can be found here: brainly.com/question/15047541

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Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

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$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

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When salt is dissolved in water, what happens to the water

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