Consider the following reaction: CrCl₃(s) + KCl(s) + 2H₂SO₄(l) $\rightarrow$KCr(SO₄)₂(s) + 4HCl(g) green white colorless purple colorless solid solid liquid solid gas When the green solid is mixed with the white solid and the colorless liquid is added, the mixture starts to bubble and fume.When all action has stopped, a wet purple solid containing solid white specks remains.Which substance is the limiting reactant?

Question

Quizplus · Accepted Answer

To determine the limiting reactant, we need to compare the amount of each reactant used in the reaction to see which one will be used up first. 
From the balanced equation, we can see that 1 mole of CrCl3 and 1 mole of KCl react with 2 moles of HCl. 
However, we don't know the amounts of each reactant present.

We can use the given information to make some assumptions: 
- The green solid (CrCl3) is one of the reactants that produce the wet purple solid, so it must have been used up in the reaction. 
- The white solid and the colorless liquid are both added to the mixture, so they must be in excess. 
- The fact that the mixture bubbles and fumes suggests that a gas (HCl) is being produced, which also indicates that the reaction is proceeding.

Therefore, the limiting reactant must be the green solid, CrCl3.

Consider the Following Reaction: CrCl3(s) + KCl(s) + 2H2SO4(l) →\rightarrow→ KCr(SO4)2(s) + 4HCl(g) Green White Colorless Purple Colorless Solid Solid

Consider the Following Reaction: CrCl₃(s) + KCl(s) + 2H₂SO₄(l) $\rightarrow$ KCr(SO₄)₂(s) + 4HCl(g) Green White Colorless Purple Colorless Solid Solid