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已知: 草酸钻

\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)

比碳酸钙受热易分解， $\mathrm{CaC}_...

Mar 19, 2024

已知: 草酸钻

\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)

比碳酸钙受热易分解，

\mathrm{CaC}_{2} \mathrm{O}_{4} \xlongequal{\text { 高温 }} \mathrm{CaCO}_{3}+\mathrm{CO} \uparrow

。将

12.8 \mathrm{gCaC}_{2} \mathrm{O}_{4}

固体加热一段时间后剩余

7.8 \mathrm{~g}

固体。下列说法正确的是 A. 剩余固体为纯净物 B. 剩余固体中含有钙元索的质量为

4.0 \mathrm{~g}

C. 将剩余固体溶于过量的稀盐酸，产生

4.4 \mathrm{gCO}_{2}

D. 加热产生的

\mathrm{CO}

气体为

2.8 \mathrm{~g}

Key Concept

Stoichiometry and mass conservation

Explanation

The mass of calcium in the remaining solid can be calculated using stoichiometry based on the molar mass of calcium and the decomposition reaction of calcium oxalate. Given that the molar mass of calcium is approximately 40 g/mol, and the initial mass of calcium oxalate is 12.8 g, we can calculate the moles of calcium oxalate and thus the moles of calcium. Since the reaction conserves the mass of calcium, the mass of calcium in the remaining solid should be the same as in the initial calcium oxalate, assuming complete decomposition of calcium oxalate to calcium carbonate. With 12.8 g of calcium oxalate, which contains 1 mole of calcium per mole of calcium oxalate, the mass of calcium is 40 g/mol * (12.8 g / 128 g/mol) = 4.0 g, where 128 g/mol is the molar mass of calcium oxalate.