The quantity of antimony in a sample can be determined by an oxidation–reduction titration with an oxidizing agent.
A 5.85 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 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is:
BrO3-(aq) + Sb3+(aq) > Br-(aq) + Sb5+(aq) (unbalanced)
(A) Calculate the amount of antimony in the sample and its percentage in the ore.

The percentage is   k  %

Explanation:

From the question we are told that

    The mass of the stibnite is  

   The volume of   KBrO3(aq) is  

     The concentration  of   KBrO3(aq) is  

Now the balanced ionic  equation for this reaction is

The number of moles of   is  

substituting values

from the reaction we see that 1 mole of  reacts with 3 moles of  

so 0.003325 moles will react with x moles of  

Therefore

Now the molar mass of is a constant with a values of  

Generally the mass of   is mathematically represented as

substituting values

The percentage of  Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as

            k  

           k  %

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2