If a solution containing 41.85 g of mercury(II) chlorate is allowed to react completely with a solution containing 12.026 g of sodium dichromate, A) how many grams of solid precipitate will be formed?
B) How many grams of the reactant in excess will remain after the reaction?
C) Assuming complete precipitation, how many moles of each ion: Hg2+, ClO3-, Na+, and Cr2(O7)2- remain in solution? (If an ion is no longer in solution, enter a zero (0) for the number of moles.)

zfoods? ok, pressure is used to keep the heat inside and to keep the food perfect.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

thats a little if u want the answer just look it up on google thats what i do haha : )