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Viefleur [7K]
3 years ago
8

For the following equation; ____S8­ + ___ O2 à ___ SO2 a. State what type of reaction is taking place? b. Balance the equation c

. If we started with 4.0 grams of Sulfur, how many moles of Sulfur Dioxide would be produced? (You must show your work)
Chemistry
1 answer:
postnew [5]3 years ago
5 0

Answer:

a) Combustion

b) S8 + 8O2  -> 8SO2

c) 0.125 mol SO2

Explanation:

a) Combustion

b) S8 + 8O2  -> 8SO2

c)

4.0 g S8 (\frac{1 mol S8}{256 g S8})(\frac{8 mol SO2}{1 mol S8} ) =0.125 mol SO2

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6.02 × 10 23

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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
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Answer:

Approximately 75%.

Explanation:

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There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

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  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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