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

Explanation:

Step1: Identify mole - ratio

From the balanced 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, if 0.020 mol of $BaCl_2$ reacts, 0.020 mol of $BaSO_4$ is formed.

Step2: Calculate mass of $BaSO_4$

Use the formula $m = n\times M$, where $n$ is the number of moles and $M$ is the molar mass. Given $n = 0.020$ mol and $M=233.39$ g/mol. Then $m=0.020\ mol\times233.39\ g/mol$. $m = 4.6678$ g.

Answer:

4.67 g (rounded to two decimal places)