# Potassium cyanide can be made in the following way: KOH (aq) + HCN (aq) ----> H2O (l) + KCN (aq) How many mL of a 1.75 M solution of HCN would be required to make 60.0 grams of potassium...

Potassium cyanide can be made in the following way: KOH (aq) + HCN (aq) ----> H2O (l) + KCN (aq) How many mL of a 1.75 M solution of HCN would be required to make 60.0 grams of potassium cyanide? Assume that there is excess KOH.

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### 2 Answers

Let's look at the equation again.

KOH + HCN --> H2O + KCN

We know that one mole of HCN is required to make one mole of KCN since both species have a coefficient of one in the above equation. We know that we want to make 60 grams of KCN. We first need to convert this to moles by dividing by the molecular weight.

60 g KCN * (1 mole/65.12 g) = 0.921 moles KCN

Since they react in a 1:1 ratio, we know that 0.921 moles of HCN are required to produce 0.921 moles of KCN. We have an aqueous solution of HCN that we know the concentration of (1.75 M), so we can calculate the volume required. M stands for molarity, or moles per liter.

0.921 moles KCN * (1 liter/1.75 moles) = 0.526 liters = 526 mL HCN

**So we need 526 mL of the HCN solution to make 60 g of KCN.**

**Sources:**

First, convert grams to moles using dimensional analysis.

`60"g" = (1 "mol")/(65.12"g") = 0.921 "mol"`

Because the balanced equation is in a 1:1 ratio, you can assume that the amount of moles KCN will equal the amount of moles HCN. Now you can use the molarity equation:

`M = "mol"/"L"`

Plug in your known values...

`1.75 "M" = (0.921"mol")/"L"`

... and solve for "L."

`"L" = (0.921"mol")/(1.75"M")`

`"L" = 0.527`

You need to convert liters to mL.

`0.527"L" = (1000"mL")/(1"L") = 527 "mL"`

If your teacher requires you to, you may need to look into sig figs (three in this case).