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A piece of rock has a mass of 2.00 g - CIE - A-Level Chemistry - Question 17 - 2016 - Paper 1

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A piece of rock has a mass of 2.00 g. It contains calcium carbonate, but no other basic substances. It neutralises exactly 36.0 cm³ of 0.500 mol dm⁻³ hydrochloric ac... show full transcript

Worked Solution & Example Answer:A piece of rock has a mass of 2.00 g - CIE - A-Level Chemistry - Question 17 - 2016 - Paper 1

Step 1

Calculate the moles of hydrochloric acid neutralized

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Answer

First, we need to calculate the total moles of hydrochloric acid used in the reaction. The formula for moles is given by:

moles=concentration×volumemoles = concentration \times volume

Using the concentration of hydrochloric acid (0.500 mol dm⁻³) and the volume in dm³ (36.0 cm³ = 0.0360 dm³):

molesHCl=0.500 mol dm3×0.0360 dm3=0.0180 molmoles_{HCl} = 0.500 \text{ mol dm}^{-3} \times 0.0360 \text{ dm}^3 = 0.0180 \text{ mol}

Step 2

Determine the moles of calcium carbonate

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Answer

The neutralization reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is:

CaCO3+2HClCaCl2+H2O+CO2CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2

From the equation, it can be seen that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid. Thus, the moles of calcium carbonate can be calculated as:

molesCaCO3=molesHCl2=0.0180 mol2=0.0090 molmoles_{CaCO_3} = \frac{moles_{HCl}}{2} = \frac{0.0180 \text{ mol}}{2} = 0.0090 \text{ mol}

Step 3

Calculate the mass of calcium carbonate

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Now that we have the moles of calcium carbonate, we can calculate its mass using the molar mass of CaCO₃, which is approximately 100.09 g/mol:

massCaCO3=molesCaCO3×molar massCaCO3=0.0090 mol×100.09 g/mol0.9009 gmass_{CaCO_3} = moles_{CaCO_3} \times molar \ mass_{CaCO_3} = 0.0090 \text{ mol} \times 100.09 \text{ g/mol} \approx 0.9009 \text{ g}

Step 4

Calculate the percentage of calcium carbonate in the rock

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Answer

Finally, we can find the percentage of calcium carbonate in the rock piece using the formula:

percentage=(massCaCO3massrock)×100=(0.9009 g2.00 g)×10045.0%percentage = \left( \frac{mass_{CaCO_3}}{mass_{rock}} \right) \times 100 = \left( \frac{0.9009 \text{ g}}{2.00 \text{ g}} \right) \times 100 \approx 45.0\%

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