Question

Calcium oxalate decomposes at 500 oC. Balance the chemical equation to find the decomposition products: CaC204(s) --> CaCO3(s) + ? What is the gaseous species that escapes? co You are correct. Your receipt no. is 152-6733 Previous Tries Calcium carbonate decomposes at 700 oC. Balance the chemical equation to find the decomposition products: CaCO3(s) --> CaO(s) + ? What is the gaseous species that escapes? Use these decomposition reactions for the following question. CO2 You are correct. Your receipt no. is 152-3670 Previous Tries A solid sample initially contains a simple mixture of calcium oxalate (CaC204) and calcium carbonate (CaCO3). This TGA graph (thermogram) plots the mass, in mg, of solid remaining as the sample temperature is raised at a constant rate, in the temperature range 350 °C to 800 °C (approx.). 9.00 (Note that CaC204 decomposes around 500 °C and CaCO3 decomposes around 700 °C.) What is the mass% of calcium carbonate in the original mixture? (Do NOT include the '%' character in your response. Try to determine your answer to 3 sig figs.) 89.19 % 8.00 Submit Answer Submission not graded. Use fewer digits. Tries 0/10 Previous Tries 7.00 Mass (mg)

Answer 1

mass % CaCO₃ = 11.1 %

Explanation

Initial mass of sample = 9.00 g

Let mass of CaC₂O₄ = x

then, mass CaCO₃ = (total mass) - (mass CaC₂O₄)

mass CaCO₃ = 9.00 g - x

moles CaC₂O₄ = (mass CaC₂O₄) / (molar mass CaC₂O₄)

moles CaC₂O₄ = (x) / (128.097 g/mol)

moles CaC₂O₄ = x/128.097 mol

moles CaCO₃ formed from decomposition of CaC₂O₄ = moles CaC₂O₄

moles CaCO₃ formed from decomposition of CaC₂O₄ = x/128.097 mol

mass CaCO₃ formed from decomposition of CaC₂O₄ = (moles CaCO₃ formed from decomposition of CaC₂O₄) * (molar mass CaCO₃)

mass CaCO₃ formed from decomposition of CaC₂O₄ = (x/128.097 mol) * (100.0869 g/mol)

mass CaCO₃ formed from decomposition of CaC₂O₄ = 0.781x g

Total mass at first point = (mass CaCO₃ formed from decomposition of CaC₂O₄) + (mass CaCO₃ in sample)

7.25 g = (0.781x g) + (9.00 g - x)

7.25 g = 0.781x g + 9.00 g - x

0.219x g = 1.75 g

x = (1.75 g) / (0.219 g)

x = 8.003 g

mass CaCO₃ = 9.00 g - x

mass CaCO₃ = 9.00 g - 8.003 g

mass CaCO₃ = 0.997 g

mass % CaCO₃ = (mass CaCO₃ / mass sample) * 100

mass % CaCO₃ = (0.997 g / 9.00 g) * 100

mass % CaCO₃ = 11.076 %