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A student completely reacts 0.403 g of magnesium oxide with an excess of carbon - OCR Gateway - GCSE Chemistry: Combined Science - Question 7 - 2021 - Paper 9

Question 7

A student completely reacts 0.403 g of magnesium oxide with an excess of carbon. 2MgO(s) + C(s) → 2Mg(s) + CO2(g) What is the mass of magnesium made? The relativ... show full transcript

Worked Solution & Example Answer:A student completely reacts 0.403 g of magnesium oxide with an excess of carbon - OCR Gateway - GCSE Chemistry: Combined Science - Question 7 - 2021 - Paper 9

Step 1

Calculate the moles of MgO

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Answer

To find the moles of magnesium oxide (MgO) that the student has, we use the formula:

$\text{Moles} = \frac{\text{mass}}{\text{molar mass}}$

The molar mass of MgO is given by:

$\text{Molar Mass of MgO} = 24.3 \text{g/mol (Mg)} + 16.0 \text{g/mol (O)} = 40.3 \text{g/mol}$

Thus, the moles of MgO are:

$\text{Moles of MgO} = \frac{0.403 \text{ g}}{40.3 \text{ g/mol}} \approx 0.0100 \text{ mol}$

Step 2

Relate moles of MgO to moles of Mg produced

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From the balanced equation:

$2 \text{MgO} (s) + \text{C}(s) \rightarrow 2 \text{Mg}(s) + \text{CO2}(g)$

we see that 2 moles of MgO produce 2 moles of Mg. Therefore, the ratio is 1:1. Thus, the moles of Mg produced will also be approximately 0.0100 mol.

Step 3

Calculate the mass of Mg produced

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Answer

Now, we can calculate the mass of magnesium produced using the formula:

$\text{Mass} = \text{Moles} \times \text{Molar Mass}$

Using the molar mass of magnesium (24.3 g/mol):

$\text{Mass of Mg} = 0.0100 \text{ mol} \times 24.3 \text{ g/mol} = 0.243 \text{ g}$

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