Question

"Смешали раствор магний сульфата с массовой долей соли 18% с раствором натрий ортофосфата массой 250гр с массовой долей соли 32,8%. Вычислить массу осадка. Помогите решить, пожалуйст

Answer 1

3MgSO4 + 2Na3PO4 = Mg3(PO4)2 (осадок)+ 3Na2SO4
n(MgSO4)= 400г*0,18%/120 г/моль =0,6 моль;
n(Na3PO4)= 250г*0,328/164 г/моль= 0,5 моль;n(MgSO4)<n(Na3PO4); расчет ведем с меньшим количеством.n(Mg3PO4)2 = 0,2 моль; 
m(Mg3(PO4)2)= 0,2 моль*262г/моль =52,4 г

Answer 2

To calculate the mass of the precipitate formed when a solution of magnesium sulfate with a mass fraction of 18% is mixed with a solution of sodium orthophosphate with a mass fraction of 32.8%, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction.

Unfortunately, I couldn't find a specific balanced chemical equation for the reaction between magnesium sulfate and sodium orthophosphate in the search results. However, I can provide you with the general steps to solve this type of problem.

Steps to solve the problem:

1. Write the balanced chemical equation for the reaction between magnesium sulfate and sodium orthophosphate, if available. 2. Determine the stoichiometry of the reaction, i.e., the mole ratio between the reactants and products. 3. Calculate the number of moles of each reactant based on their mass and mass fraction. 4. Use the stoichiometry to determine the number of moles of the precipitate formed. 5. Convert the moles of the precipitate to its mass using its molar mass.

Since I couldn't find the specific balanced chemical equation, I won't be able to provide you with the exact calculation. However, I can guide you through the general steps mentioned above if you have the balanced chemical equation for the reaction between magnesium sulfate and sodium orthophosphate.