For each precipitation reaction, calculate how many grams of the first reactant are necessary to completely react with 56.5 g of the second reactant. k2so4(aq)+sr(no3)2(aq)→srso4(s)+2kn o3(aq)

The precipitation reaction is as follows:

Here, 1 mole of reactant 1 reacts with 1 mole of reactant 2 .

The mass of is 56.5 g and molar mass is 211.63 g/mol, converting mass into molar mass as follows:

Since, 1 mol of reacts with 1 mol of  thus, 0.267 mol will react with 0.267 mol.

Molar mass of is 174.259 g/mol, mass of  can be calculated as follows:

Therefore, mass of first reactant is 46.52 g

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"