ti= 3.0g/ 204.3833 =1.47 x 10^-2
cl2= 8.0/35.45 /2=1.1 x 10^-1
ti is the Lr since it produces less product.
Usng the following balanced chemical equation, determine the limiting reactant in the reaction between 3.0 grams of titanium and 8.0 grams of chlorine gas?
Ti(s) + 2Cl2(g) --> TiCl4(s)
2 answers
Moles of Ti = 3.0g / 47.87g/mol = 0.0627 moles Ti
8.0g Cl2 / 70.9g/mol = 0.113 mol Cl2
Available ratio:
(0.113 mol Cl2) / (0.0627 moles Ti) =
1.80 mol Cl2 / 1mole Ti
Needed Ratio:
The Cl2 / Ti mole ratio in the balanced chemical equation =
2 mol Cl2 / 1 mol Ti
The comparison of the mole ratio in the mixture to the mole ratio needed shows we don't have enough Cl2 to react with all the Ti metal. So, Cl2 is the Limiting Reagent, and Ti is used in excess.
8.0g Cl2 / 70.9g/mol = 0.113 mol Cl2
Available ratio:
(0.113 mol Cl2) / (0.0627 moles Ti) =
1.80 mol Cl2 / 1mole Ti
Needed Ratio:
The Cl2 / Ti mole ratio in the balanced chemical equation =
2 mol Cl2 / 1 mol Ti
The comparison of the mole ratio in the mixture to the mole ratio needed shows we don't have enough Cl2 to react with all the Ti metal. So, Cl2 is the Limiting Reagent, and Ti is used in excess.
3 answers