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Imagine you conduct a decomposition experiment of magnesium carbonate, MgCO3. If 25.0g of MgCO3 is heated to form 9.7 g MgO, what is the percent yield

User Ducarmel
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To calculate the percent yield of the decomposition of magnesium carbonate (MgCO3), you need to compare the actual yield (the amount of product obtained experimentally) to the theoretical yield (the maximum amount of product that could be obtained according to stoichiometry).

The balanced chemical equation for the decomposition of magnesium carbonate is:

MgCO3(s) → MgO(s) + CO2(g)

First, calculate the molar mass of MgCO3 and MgO:

MgCO3:

- Mg: 24.31 g/mol

- C: 12.01 g/mol

- O: 16.00 g/mol

MgCO3 molar mass = 24.31 + 12.01 + (3 × 16.00) = 84.32 g/mol

MgO:

- Mg: 24.31 g/mol

- O: 16.00 g/mol

MgO molar mass = 24.31 + 16.00 = 40.31 g/mol

Now, calculate the theoretical yield of MgO by using stoichiometry. Start with the given mass of MgCO3 (25.0 g) and convert it to moles using the molar mass of MgCO3:

\[25.0 g \text{ MgCO3} \left(\frac{1 \text{ mol MgCO3}}{84.32 \text{ g MgCO3}}\right) = 0.297 mol MgCO3\]

According to the balanced equation, 1 mole of MgCO3 produces 1 mole of MgO, so you have 0.297 moles of MgO as the theoretical yield.

Now, convert this to grams using the molar mass of MgO:

\[0.297 mol MgO \left(\frac{40.31 g MgO}{1 \text{ mol MgO}}\right) = 11.98 g MgO\]

This is the theoretical yield of MgO.

Now, calculate the percent yield using the actual yield (9.7 g) and the theoretical yield (11.98 g):

\[\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\]

\[\text{Percent Yield} = \frac{9.7 g}{11.98 g} \times 100 = 81.17\%\]

The percent yield of the decomposition of magnesium carbonate is approximately 81.17%.

User Bavan
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