110.4 g of dolomite was calcined, as a result of which a gas was released, which contains 18.06 × 10²²

| May 26, 2021 | education

110.4 g of dolomite was calcined, as a result of which a gas was released, which contains 18.06 × 10²² molecules, determine the mass and composition of the resulting residue.

CaMg (CO3) 2 = CaO + MgO + 2CO2. CO2 amount: n = N / Na = 18.06 * 10 ^ 22 / 6.02 * 10 ^ 23 = 0.3 mol. It is 2 times more than the amount of CaO and MgO, which means n (CaO) = n (MgO) = 0.3 / 2 = 0.15 mol. m (CaO) = 0.15 * 56 = 8.4g; m (MgO) = 0.15 * 40 = 6g. The amount of СaMg (CO3) 2: n = m / M = 110.4 / 184 = 0.6 mol. According to the equation, the amounts of CaMg (CO3) 2, CaO and MgO are equal (i.e. 0.15 mol). It turns out that 0.45 mol of CaMg (CO3) 2 did not react (0.6-0.15 = 0.45). m (sediment) = m (CaO) + m (MgO) + m (unreacted CaMg (CO3) 2) = 8.4 + 6 + 0.45 * 184 = 97.2g. Sediment composition: CaO, MgO, CaMg (CO3) 2.



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