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Metal oxides are produced when metals are heated in air - AQA - GCSE Chemistry Combined Science - Question 6 - 2021 - Paper 1
Question 6
Metal oxides are produced when metals are heated in air. A student investigated the change in mass when 0.12 g of magnesium was heated in air. Figure 5 shows the a... show full transcript
Worked Solution & Example Answer:Metal oxides are produced when metals are heated in air - AQA - GCSE Chemistry Combined Science - Question 6 - 2021 - Paper 1
Step 1
Calculate the number of moles of oxygen gas (O2) that reacted.
Answer
To calculate the number of moles of oxygen gas that reacted, we first determine the mass of oxygen in the reaction:
Mass of oxygen = Mass of magnesium oxide - Mass of magnesium
Substituting the values:
Mass of oxygen = 0.20 g - 0.12 g = 0.08 g
Next, we convert the mass of oxygen into moles using the formula:
Number of moles = ( \frac{\text{mass}}{\text{molar mass}} )
The molar mass of O2 is ( 32 , g/mol ). Therefore:
[ \text{Number of moles of O2} = \frac{0.08 , g}{32 , g/mol} = 0.0025 , moles ]
Step 2
Suggest why the mass of magnesium oxide produced would be different without a lid on the crucible.
Answer
Without a lid, the mass of magnesium oxide produced would be less because some of the magnesium oxide could escape into the air as vapor or could be lost due to incomplete reactions or environmental factors.
Step 3
Calculate the mass of copper oxide produced when 0.5 g of copper reacts with oxygen.
Answer
To find the mass of copper oxide produced, we use the ratio of copper to copper oxide from the given data:
[ \frac{63.5 , g : ext{Cu}}{79.5 , g : ext{CuO}} = \frac{0.5 , g : ext{Cu}}{x , g : ext{CuO}} ]
Cross-multiplying gives:
[ 63.5 , g : ext{Cu} \cdot x = 79.5 , g \cdot 0.5 , g ]
Solving for ( x ):
[ x = \frac{79.5 \cdot 0.5}{63.5} = 0.6259 , g \approx 0.626 , g : (3 , ext{significant figures}) ]
Step 4
Determine the formula of the iron oxide produced.
Answer
Given that 0.015 moles of iron reacts with 0.010 moles of oxygen, we can determine the simplest whole number ratio:
[ \text{Molar ratio of Fe: O2} = \frac{0.015}{0.010} = 1.5:1 ]
To find whole numbers, we multiply both by 2:
[ 3:2 \rightarrow \text{Formula of iron oxide} = \text{Fe}_3\text{O}_4 ]
Step 5