1. A solution containing 0.100 mol of Na2CO3 and 0.100 mol of NiCl2 is allowed to react. Which reactant is the limiting reactant?
2. A solution containing 0.100 mol of Na2CO3 and 0.100 mol of NiCl2 is allowed to react. How many moles of precipitate form in this reaction?
Limiting reagent problems are worked as a special case of regular stoichiometry problems. Here are two example problems, both worked completely. Just follow the steps.
If you have trouble, post your work ad explain in detail the trouble you are having and I'll be happy to help you through it.
stoichiometry problem here.
http://www.jiskha.com/science/chemistry/stoichiometry.html
limiting reagent problem here.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
Can you explain how I get the moles and I should be able to do it from there.
Both problems give you the mols to start.
You CONVERT mols of what you are given to mols of the product by using the coefficients in the balanced equation. For example,
Na2CO3 + NiCl2 ==> NiCO3 + 2NaCl
You have 0.1 mol Na2CO3. How many mols NiCO3 will it form if you have 0.1 mo Na2CO3 and all of the NiCl2 you need. You will obtain 01 mol Na2CO3 x (1 mol NiCO3/1 mol Na2CO3) = 0.1 mol x (1/1) = 0.1 mol NiCO3.
Thank You that really helped
To determine the limiting reactant in a chemical reaction, you need to compare the moles of each reactant and their respective stoichiometric coefficients.
1. Begin by writing the balanced chemical equation for the reaction between Na2CO3 and NiCl2:
Na2CO3 + NiCl2 -> NiCO3 + 2NaCl
2. Next, calculate the number of moles of the reactants in the given solution:
- 0.100 mol of Na2CO3
- 0.100 mol of NiCl2
3. To find the limiting reactant, compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. The stoichiometric coefficient of Na2CO3 is 1, while the stoichiometric coefficient of NiCl2 is also 1.
4. Since the moles of both reactants are equal, neither reactant is in excess, and both will be completely consumed in the reaction. Hence, in this case, there is no limiting reactant.
Moving on to the second question:
To find the number of moles of precipitate formed in the reaction, you need to determine the stoichiometry of the reaction and the ratio of the precipitate to the reactants.
1. Refer to the balanced chemical equation:
Na2CO3 + NiCl2 -> NiCO3 + 2NaCl
2. The stoichiometric coefficient of NiCO3 in the balanced equation is 1, indicating that one mole of NiCO3 is produced for every mole of the limiting reactant.
3. In this case, since there is no limiting reactant (as determined in the previous question), the amount of precipitate formed will depend on the full reaction of both reactants.
4. As both Na2CO3 and NiCl2 have equal moles (0.100 mol), the reaction will form 0.100 mol of NiCO3.
Therefore, 0.100 mol of precipitate (NiCO3) will form in this reaction when a solution containing 0.100 mol of Na2CO3 and 0.100 mol of NiCl2 is allowed to react.