In the reaction of formation of sodium chloride (NaCl), 15 g of sodium is placed in contact with 20 g of chlorine. What is the reactant in excess and
its mass in excess?
2Na(s) + Cl2(g) => 2NaCl(s)
Convert given data to moles and divide each by the respective coefficient in the balanced equation. The smaller value is the limiting reagent and the larger value is the one that will remain in excess.
moles Na = (15g/23g-mol^-1)mole = 0.652 mole Na
moles Cl2 = (20g/70g-mol^-1)mole = 0.286 mole Cl2
Na => (0.652/2)=0.326
Cl2 => (0.286/1)= 0.286
Cl2 is the Limiting reagent and Na will remain in excess upon completion of the reactions.
Thanks! =)
To determine the reactant in excess and its mass, we need to compare the stoichiometry of the reaction to the given amounts of sodium and chlorine.
The balanced equation for the formation of sodium chloride is as follows:
2 Na + Cl2 -> 2 NaCl
According to the equation, it takes 2 moles of sodium (Na) to react with 1 mole of chlorine gas (Cl2) to form 2 moles of sodium chloride (NaCl).
To find the limiting reactant, we need to calculate the moles of each reactant. To do this, we can use the molar mass:
- Molar mass of sodium (Na) = 23 g/mol
- Molar mass of chlorine (Cl2) = 35.5 g/mol
Now, let's calculate the moles of each reactant:
Moles of sodium (Na) = Mass of sodium (g) / Molar mass of sodium (g/mol)
Moles of sodium (Na) = 15 g / 23 g/mol = 0.6522 mol
Moles of chlorine (Cl2) = Mass of chlorine (g) / Molar mass of chlorine (g/mol)
Moles of chlorine (Cl2) = 20 g / 35.5 g/mol = 0.5634 mol
Now, we can compare the moles of each reactant to the stoichiometry of the reaction.
According to the balanced equation, the ratio of moles of sodium (Na) to moles of chlorine (Cl2) is 2:1. This means that for every 2 moles of sodium, we need 1 mole of chlorine gas.
In this case, since we have fewer moles of chlorine (0.5634 mol) compared to sodium (0.6522 mol), chlorine is the limiting reactant. This indicates that sodium is in excess.
To determine the mass of the reactant in excess (sodium in this case), we need to calculate the remaining mass of sodium that did not react.
Mass of sodium in excess (g) = Total mass of sodium (g) - Mass of sodium used in the reaction (g)
Mass of sodium in excess (g) = 15 g - (moles of sodium used * Molar mass of sodium)
Mass of sodium in excess (g) = 15 g - (0.5634 mol * 23 g/mol) = 2.3802 g
Therefore, the reactant in excess is sodium, and the mass of sodium in excess is 2.3802 g.