Question

1.)

If a solution containing 56.59 g56.59 g of mercury(II) nitrate is allowed to react completely with a solution containing 17.796 g17.796 g of sodium sulfate according to the equation below.

Hg(NO3)2(aq)+Na2SO4(aq)⟶2NaNO3(aq)+HgSO4(s)Hg(NO3)2(aq)+Na2SO4(aq)⟶2NaNO3(aq)+HgSO4(s)

How many grams of solid precipitate will be formed?

How many grams of the reactant in excess will remain after the reaction?

2.)

Each step in the following process has a yield of 60.0%.60.0%.

CH4+4Cl2⟶CCl4+4HClCH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HClCCl4+2HF⟶CCl2F2+2HCl

The CCl4CCl4 formed in the first step is used as a reactant in the second step.

If 4.00 mol4.00 mol CH4CH4 reacts, what is the total amount of HClHCl produced? Assume that Cl2Cl2 and HFHF are present in excess. Moles of HCl?

3.):

Cryolite, Na3AlF6(s),Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.

Balance the equation for the synthesis of cryolite.    Al2O3 (s) +NaOH(l)+ HF(g)------->Na3AlF6+H2O (g)

If 17.2 kg of Al2O3(s),17.2 kg of Al2O3(s), 56.4 kg of NaOH(l),56.4 kg of NaOH(l), and 56.4 kg of HF(g)56.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

mass of cryolite produced:?

Which reactants will be in excess?

What is the total mass of the reactants left over after the reaction is complete? kg?

Answer 1

Dear Student, Kindly refer to the solution of your first question.