Write the balanced complete equation for the reaction, and then write the net io

Question

Write the balanced complete equation for the reaction, and then write the net ionic equation. Do not forget state symbols? A reaction initially contains 0.233 moles of chromium (II) sulfide and 0.652 moles of hydrochloride acid. When the reaction is complete, what reagent is left over, and how much of that reagent is left (in moles)? The chromium-containing product in the reaction has a molar mass of 122.8. Assume I run the reaction as described in (b) above, and I find that my percent chromium-containing product is 78%. What about (in grams) of chromium product did I get?

Explanation / Answer

Balanced chemical reaction of CrS and HCl

CrS(s) +2 HCl (aq) -- > CrCl2 (aq) + H2S (g)

Net ionic equation

To write net ionic equation we split aqueous terms into ions.

CrS(s) +2 H+ + 2Cl- (aq) -- >2 Cr2+    + 2 Cl- (aq) + H2S (g)

Since Cl- ions are present to both side of the equation so we can cancel them. These ions are called as spectator ions.

Then equation becomes.

CrS(s) +2 H+ (aq)   -- >2 Cr2+ ( aq) + H2S (g)

b).

Given : CrS =0.233

Mol of HCl = 0.652

Calculation of limiting reactant

1 mol CrS needs 2 mol HCl

Moles of HCl needed to react with 0.233 mol CrS

= 0.233 mol CrS x 2 mol HCl / 1 mol CrS

=0.466 mol HCl

Actually there is 0.652 moles of HCl is present so CrS is limiting reactant.

Now we calculate excess reactant

n HCl = Original moles - moles of HCl reacted

= 0.652 – 0.466 = 0.186 mol

Moles of HCl remained in the solution = 0.186

3 ) To get actual amount of product we need to get theoretical yield.

Theoretical yield we get from moles of product formed from limiting reactant.

Moles of CrCl2 = moles of CrS x 1 mol CrCl2 / 1 mol CrS

= 0.233 of CrCl2

Mass of CrCl2 = mol x molar mass

= 0.233 mol x 122.8 g /mol

= 28.61 g

This is theoretical yield of CrCl2

Percent yield = (actual yield / theoretical yild ) x 100

We are given percent yield

78 % = actual yield / 28.61 g ) x 100

Actual yield = 22.32 g

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