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.
ScanSolve AI Answer
Here's how to calculate the amount of copper(II) oxide (CuO) that will react with 4.9 \, g of tetraoxosulphate(VI) acid (H_2SO_4): Step 1: Write the balanced chemical equation. The reaction between tetraoxosulphate(VI) acid (H_2SO_4) and copper(II) oxide (CuO) is: H_2SO_4 + CuO → CuSO_4 + H_2O This equation is already balanced, showing a 1:1 mole ratio between H_2SO_4 and CuO. Step 2: Calculate the molar masses of the reactants. Given atomic masses: Cu = 64, O = 16, S = 32, H = 1. Molar mass of H_2SO_4: (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 \, g mol^-1 Molar mass of CuO: 64 + 16 = 80 \, g mol^-1 Step 3: Use stoichiometry to find the mass of CuO. From the balanced equation, 1 mole of H_2SO_4 reacts with 1 mole of CuO. This means 98 \, g of H_2SO_4 reacts with 80 \, g of CuO. We can set up a proportion to find the mass of CuO (x) that reacts with 4.9 \, g of H_2SO_4: Mass of H_2SO_4Molar mass of H_2SO_4 = Mass of CuOMolar mass of CuO 4.9 \, g98 \, g mol^-1 = x \, g80 \, g mol^-1 Now, solve for x: x = 4.9 \, g98 \, g mol^-1 × 80 \, g mol^-1 x = 0.05 \, mol × 80 \, g mol^-1 x = 4.0 \, g The amount of copper(II) oxide that will react is 4.0 \, g. The final answer is D.