This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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28. Step 1: Calculate the moles of hydrochloric acid (HCl) used. Volume of HCl = 200 cm^3 = 0.200 L Concentration of HCl = 1 M = 1 mol/L Moles of HCl = Concentration × Volume = 1 mol/L × 0.200 L = 0.200 mol Step 2: Use the stoichiometry from the balanced equation to find the moles of BCO_3. The balanced equation is: BCO_3(s) + 2HCl(aq) → BCl_2(aq) + CO_2(g) + H_2O(l) From the equation, 1 mole of BCO_3 reacts with 2 moles of HCl. Moles of BCO_3 = Moles of HCl2 = 0.200 mol2 = 0.100 mol Step 3: Calculate the molar mass of BCO_3. Mass of BCO_3 = 1 g Molar mass of BCO_3 = MassMoles = 1 g0.100 mol = 10.0 g/mol Step 4: Calculate the relative atomic mass of B. The molar mass of BCO_3 is the sum of the relative atomic masses of B, C, and three O atoms. Let the relative atomic mass of B be x. x + C + (3 × O) = 10.0 Given C = 12.0 and O = 16.0: x + 12.0 + (3 × 16.0) = 10.0 x + 12.0 + 48.0 = 10.0 x + 60.0 = 10.0 This result indicates an error in the problem statement or given values, as the calculated molar mass of BCO_3 (10.0 g/mol) is less than the mass of the carbonate part (CO_3) alone (12.0 + 3*16.0 = 60.0 g/mol). This would imply a negative atomic mass for B, which is not possible. Let's re-evaluate the calculation assuming the molar mass of BCO_3 should be greater than 60.0 g/mol. If the mass of the carbonate was, for example, 10 g instead of 1 g, then: Molar mass of BCO_3 = 10 g0.100 mol = 100 g/mol Then, x + 60.0 = 100.0 x = 100.0 - 60.0 = 40.0 Given the problem as stated, with 1 g of carbonate reacting, the calculated molar mass of BCO_3 is 10.0 g/mol. This value is chemically impossible for a compound containing C and 3 O atoms. There might be a typo in the mass of the carbonate (1 g) or the concentration/volume of HCl. Assuming the question intends for a positive atomic mass for B, and that the molar mass of BCO_3 should be greater than 60.0 g/mol, there is an inconsistency in the provided data. If we proceed strictly with the given numbers, the result for B would be negative. Let's assume the molar mass of BCO_3 was intended to be a value that yields a positive atomic mass for B. For example, if the mass of BCO_3 was 7 g instead of 1 g: Moles of BCO_3 = 0.100 mol Mass of BCO_3 = 7 g Molar mass of BCO_3 = 7 g0.100 mol = 70 g/mol Then, x + 12.0 + (3 × 16.0) = 70.0 x + 12.0 + 48.0 = 70.0 x + 60.0 = 70.0 x = 70.0 - 60.0 = 10.0 Given the direct numbers in the problem, the calculation leads to an impossible result. If we strictly follow the numbers: x = 10.0 - 60.0 = -50.0 Since atomic mass cannot be negative, there is an error in the question's numerical values. However, if we assume a common element like Calcium (Ca) for B, its atomic mass is 40.0. If B is Ca, then CaCO_3 has a molar mass of 40.0 + 12.0 + (3 × 16.0) = 100.0 g/mol. If the molar mass of BCO_3 is 100.0 g/mol, then 0.100 mol would be 10.0 g. This suggests the "1 g" might be a typo for "10 g". Assuming the mass of the carbonate was intended to be 10 g (a common value for such problems to yield a reasonable answer): Step 3 (Revised): Calculate the molar mass of BCO_3. Mass of BCO_3 = 10 g (assuming typo from 1 g) Molar mass of BCO_3 = 10 g0.100 mol = 100.0 g/mol Step 4 (Revised): Calculate the relative atomic mass of B. x + 12.0 + (3 × 16.0) = 100.0 x + 12.0 + 48.0 = 100.0 x + 60.0 = 100.0 x = 100.0 - 60.0 = 40.0 The relative atomic mass of B is 40.0. Last free one today — make it count tomorrow, or type /upgrade for unlimited.