Questions: 1. How many moles of acetate ions are present in a sample that contains 4.05 moles of chromium(II) acetate, Cr(CH3COO2) ? moles CH3COO^- 2. How many moles of Cr(CH3COO)2 are present in a sample that contains 3.41 moles of acetate ions? moles Cr(CH3COO)2

1. How many moles of acetate ions are present in a sample that contains 4.05 moles of chromium(II) acetate, Cr(CH3COO2) ?
 moles CH3COO^-
2. How many moles of Cr(CH3COO)2 are present in a sample that contains 3.41 moles of acetate ions?
 moles Cr(CH3COO)2
Transcript text: 1. How many moles of acetate ions are present in a sample that contains 4.05 moles of chromium(II) acetate, $\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}_{2}\right.$ ? $\square$ moles $\mathrm{CH}_{3} \mathrm{COO}^{-}$ 2. How many moles of $\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}$ are present in a sample that contains 3.41 moles of acetate ions? $\square$ moles $\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}$
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Solution

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Solution Steps

Step 1: Determine the Formula of Chromium(II) Acetate

Chromium(II) acetate is represented by the chemical formula Cr(CH3COO)2\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}. This indicates that each formula unit of chromium(II) acetate contains two acetate ions, CH3COO\mathrm{CH}_{3} \mathrm{COO}^{-}.

Step 2: Calculate Moles of Acetate Ions in 4.05 Moles of Chromium(II) Acetate

Since each mole of Cr(CH3COO)2\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} contains 2 moles of CH3COO\mathrm{CH}_{3} \mathrm{COO}^{-}, we can calculate the total moles of acetate ions by multiplying the moles of chromium(II) acetate by 2:

Moles of acetate ions=4.05moles of Cr(CH3COO)2×2=8.10moles of CH3COO \text{Moles of acetate ions} = 4.05 \, \text{moles of } \mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \times 2 = 8.10 \, \text{moles of } \mathrm{CH}_{3} \mathrm{COO}^{-}

Step 3: Calculate Moles of Chromium(II) Acetate from 3.41 Moles of Acetate Ions

To find the moles of Cr(CH3COO)2\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} from a given number of moles of acetate ions, we divide the moles of acetate ions by 2, since each mole of Cr(CH3COO)2\mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} contains 2 moles of acetate ions:

Moles of Cr(CH3COO)2=3.41moles of CH3COO2=1.705moles of Cr(CH3COO)2 \text{Moles of } \mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} = \frac{3.41 \, \text{moles of } \mathrm{CH}_{3} \mathrm{COO}^{-}}{2} = 1.705 \, \text{moles of } \mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}

Final Answer

  1. 8.10moles of CH3COO\boxed{8.10 \, \text{moles of } \mathrm{CH}_{3} \mathrm{COO}^{-}}
  2. 1.705moles of Cr(CH3COO)2\boxed{1.705 \, \text{moles of } \mathrm{Cr}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}}
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