we wish to determine the mass of baso₄ formed when 200. ml of 0.10 m bacl₂ reacts with excess na₂so₄…

we wish to determine the mass of baso₄ formed when 200. ml of 0.10 m bacl₂ reacts with excess na₂so₄ according to the equation below. bacl₂(aq) + na₂so₄(aq) → baso₄(s) + 2nacl(aq) in the previous step, you determined 0.020 mol bacl₂ reacts. the molar mass of baso₄ is 233.39 g/mol. what mass of baso₄ forms during the reaction?

we wish to determine the mass of baso₄ formed when 200. ml of 0.10 m bacl₂ reacts with excess na₂so₄ according to the equation below. bacl₂(aq) + na₂so₄(aq) → baso₄(s) + 2nacl(aq) in the previous step, you determined 0.020 mol bacl₂ reacts. the molar mass of baso₄ is 233.39 g/mol. what mass of baso₄ forms during the reaction?

Answer

Answer:

4.678 g

Explanation:

Step1: Calculate moles of BaCl₂

The volume of BaCl₂ is (V = 200\ mL=0.200\ L) and the molarity (M = 0.10\ M). Using the formula (n = M\times V), we have (n_{BaCl_2}=0.10\ mol/L\times0.200\ L = 0.020\ mol).

Step2: Determine mole - ratio

From the balanced chemical equation (BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)), the mole - ratio of (BaCl_2) to (BaSO_4) is (1:1). So, (n_{BaSO_4}=n_{BaCl_2}=0.020\ mol).

Step3: Calculate mass of BaSO₄

The molar mass of (BaSO_4) is (M_{BaSO_4}=233.39\ g/mol). Using the formula (m = n\times M), we get (m_{BaSO_4}=0.020\ mol\times233.39\ g/mol = 4.6678\ g\approx4.678\ g).