How many grams are in 1.11 moles of manganese sulfate ?

Step 1: Find the molar mass of Mn3(SO4)7.  

Mn3(SO4)7 = 3(54.94) + 7(32.07) + 28(16) = 837.31g  

Step 2: Calculate grams of Mn3(SO4)7 in the following manner:  

1.11 mol Mn3(SO4)7 (837.31g Mn3(SO4)7 /1 mol Mn3(SO4)7 ) = 929g Mn3(SO4)7  

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.