Question

Question #6c3f7

Answer 1

`"509 mL"`

Explanation:

The problem provides you with the balanced chemical equation for this neutralization reaction

`"KOH"_text((aq]) + "HCN"_text((aq]) -> "H"_2"O"_text((l]) + "KCN"_text((aq])`

Hydrocyanic acid, which is the name given to hydrogen cyanide, `"HCN"`, when dissolved in aqueous solution, will react with potassium hydroxide, `"KOH"`, to form water and aqueous potassium cyanide, `"KCN"`.

Notice that one mole of hydrocyanic acid will produce one mole of potassium cyanide. This means that if you can determine the number of moles of potassium cyanide produced by the reaction, you can backtrack to find the number of moles of hydrocyanic acid that took part in the reaction.

Use potassium cyanide's molar mass to determine how many moles were produced by the reaction

`50.0 color(red)(cancel(color(black)("g"))) * "1 mole KCN"/(65.116color(red)(cancel(color(black)("g")))) = "0.7679 moles KCN"`

Since potassium hydroxide is in excess, it will not limit the number of moles of hydrocyanic acid that can take part in the reaction.

According to the aforementioned `1:1` mole ratio, if `0.7679` moles of potassium cyanide were produced by the reaction, then `0.7679` moles of hydrocyanic acid must have reacted with the potassium hydroxide.

Now, the problem provides you with the molarity of the hydrocyanic acid solution. Molarity tells you the number of moles of solute present in one liter of solution

`color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))`

In your case, a molarity of `"1.51 M"` tells you that one mole of this solution contains `1.51` moles of hydrocyanic acid

`"1.51 M" = "1.51 mol L"^(-1)`

You can thus use the molarity of the solution as a conversion factor to determine how many liters of this solution would contain `0.7679` moles

`0.7679color(red)(cancel(color(black)("moles HCN"))) * overbrace("1 L solution"/(1.51color(red)(cancel(color(black)("moles HCN")))))^(color(purple)("1.51 M")) = "0.50854 L"`

Rounded to three sig figs and expressed in milliliters, the answer will be

`"volume" = color(green)(|bar(ul(color(white)(a/a)"509 mL"color(white)(a/a)|)))`

Keep in mind that

`"1 L" = 10^3"mL"`