Could you please answer these questions 1- 2- 3- 4- 5- 6- Thank you If a solutio
ID: 829262 • Letter: C
Question
Could you please answer these questions
1-
2-
3-
4-
5-
6-
Thank you
If a solution containing 46.661 g of mercury(ll) perchlorate is allowed to react completely with a solution containing 13.180 g of sodium sulfate, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction? The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.33-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+(aq). The Sb3 + (aq) is completely oxidized by 26.6 mL of a 0.135 M aqueous solution of KBrQ3 (aq). The unbalanced equation for the reaction is Calculate the amount of antimony in the sample and its percentage in the ore. The amount of I3 (aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32-(aq) (thiosulfate ion). The determination is based on the net ionic equation Complete combustion of 2.20 g of a hydrocarbon produced 6.66 g of CO2 and 3.41 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary An unknown compound contains only C, H, and O. Combustion of 6.30 g of this compound produced 15.4 g of CO2 and 6.30 g of H2O. What is the empirical formula of the unknown compound? Insert subscripts as needed. Calculate the molarity of the following solutions, 24.9 g of NaCI in 663 mL of solution. 0.550 mol of NaOH in 1.20 L of solution.Explanation / Answer
1).
Hg(ClO4)2(aq) + Na2SO4(aq) ---> HgSO4(s) + 2NaClO4 (aq)
1 mol Mercury(II) Perchlorate per 2 mol Na2SO4
Mol HgClO4 = 46.661 g / 399.49 g/mol = 0.117
Mol Na2SO4 = 13.18 g / 142.04 g/mol = 0.093
So sodium sulfate is the limiting reagant. The reaction will yeild: 0.093 mol of precipitate
0.093 mol * 296.65 g/mol = 27.6 grams of precipitate (Mercury Sulfate)
There will be 0.117 - 0.093 = 0.024 mol of reactant left in solution
0.024*399.49 = 9.59 grams of reactant left in solution
2). Balanced equation: 6H+(aq) + BrO3