Photo AI

Metal oxides are produced when metals are heated in air - AQA - GCSE Chemistry Combined Science - Question 6 - 2021 - Paper 1

Question icon

Question 6

Metal-oxides-are-produced-when-metals-are-heated-in-air-AQA-GCSE Chemistry Combined Science-Question 6-2021-Paper 1.png

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

0.12 g of magnesium reacted to produce 0.20 g of magnesium oxide. Calculate the number of moles of oxygen gas (O₂) that reacted.

96%

114 rated

Answer

To find the mass of oxygen that reacted, we first subtract the mass of magnesium from the mass of magnesium oxide:

extMassofoxygen=extMassofmagnesiumoxideextMassofmagnesium=0.20extg0.12extg=0.08extg ext{Mass of oxygen} = ext{Mass of magnesium oxide} - ext{Mass of magnesium} = 0.20 ext{ g} - 0.12 ext{ g} = 0.08 ext{ g}

Next, we calculate the number of moles of oxygen gas using its relative atomic mass:

ext{Moles of oxygen (O}_2 ext{)} = rac{ ext{Mass of oxygen}}{ ext{Relative atomic mass of O}_2} = rac{0.08 ext{ g}}{32} = 0.0025 ext{ moles}

Step 2

The student repeated the experiment without a lid on the crucible. Suggest why the mass of magnesium oxide produced would be different without a lid on the crucible.

99%

104 rated

Answer

Without a lid on the crucible, magnesium oxide can escape during the reaction. This would result in a lower mass of magnesium oxide being produced because some of the product is lost to the environment.

Step 3

63.5 g of copper produces 79.5 g of copper oxide. Calculate the mass of copper oxide produced when 0.50 g of copper reacts with oxygen.

96%

101 rated

Answer

We start by establishing a ratio from the given data:

ext{Mass of copper oxide} = rac{79.5 ext{ g}}{63.5 ext{ g}} imes 0.50 ext{ g} = 0.6259 ext{ g}

Rounding this to three significant figures gives:

extMassofcopperoxide=0.626extg ext{Mass of copper oxide} = 0.626 ext{ g}

Step 4

0.015 moles of iron reacts with 0.010 moles of oxygen (O₂). Determine the formula of the iron oxide produced and the balanced symbol equation for the reaction.

98%

120 rated

Answer

The reaction of iron with oxygen to form iron oxide can be framed as follows:

  1. From the mole ratio of iron to oxygen:
ightarrow ext{Fe}_2 ext{O}_3$$ To find the ratio: $$ 4 ext{ Fe} + 3 ext{ O}_2 ightarrow 2 ext{ Fe}_2 ext{O}_3.$$ 2. The balanced equation is: $$3 ext{ Fe} + ext{ O}_2 ightarrow ext{Fe}_3 ext{O}_4$$ This means the formula of the iron oxide produced is $ ext{Fe}_3 ext{O}_4$.

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;