For the following chemical reaction, given 2.5 gram mgso4 with an excess of ba(no3)2, calculate the theoretical mass of baso4 precipitate produced from the reaction (in grams). the following conceptual plan is designed to you convert the mass of the limiting reactant (mgso4) to the mass of the desired product (baso4).

The answer to your question is:  4.9 g of BaSO4

Explanation:

Data

MgSO4 = 2.5 g

Ba(NO3)2 = excess reactant

BaSO4 = ?

Reaction

                  MgSO4   +    Ba(NO3)2      ⇒     BaSO4   +    Mg(NO3)2

MW MgSO4 = 24 + 32 + 64

                    = 120 g

MW BaSO4 = 137 + 32 + 64

                    = 233 g

                  120 g of MgSO4  233g of BaSO4

                  2.5 g                       x

                  x = 4.9 g of BaSO4

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.