# 4.9 - pretty easy to do if we know the charges on each...

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Key for Micro Exam 4 Question 9 Question 9: (Try this problem without a calculator) How many moles of cyanide ions are found in 13.0 g of chromium(III) cyanide? Answers: (a) 0.10 mol (b) 0.20 mol (c) 0.30 mol (d) 0.40 mol (e) 0.50 mol (f) 1.0 mol (g) 3.0 mol Any time you are converting from mass to moles you should think dimensional analysis. A scheme would looks something like this: g chromium(III) cyanide → moles chromium(III) cyanide → moles cyanide We can’t do much with just the written-out name of the ionic compound because it only tells us about charge, not about how many of each atom are in the ionic compound. We should figure out the chemical formula which does tell us the number of each molecule, and luckily this is
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Unformatted text preview: pretty easy to do if we know the charges on each ionic participant. The Roman numeral after chromium tells us that chromium, when separated from cyanide, would have a +3 charge. You should have memorized that cyanide ion (CN-) has a -1 charge. Ionic compounds are neutral, so we need to add enough chromium ions and cyanide ions to make a neutral compound. This would occur with one chromium cation and three cyanide anions like this Cr +3 + 3CN-1 → Cr(CN) 3 Oftentimes the charge on one ion will become the subscript on the other ion when written as an ionic compound. Now let’s do the dimensional analysis: 13g Cr(CN) 3 │ 1mol Cr(CN) 3 │ 3mol CN-= 0.3g CN-│ 130g Cr(CN) 3 │ 1 mol Cr(CN) 3...
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