A student wants to conduct an experiment to see how much NaCl can be produced in the reaction between HCl and NaOH with water as the additional product. The student has 3.21 g of NaOH, if the student adds excess HCl to the NaOH how much NaCl will be produced? Show all your work.
3.21 g NaOH = 3.21/40 = 0.08 moles
NaOH + HCl = NaCl + H2O
so you will get that many moles of NaCl and H2O
convert back to grams
Well, let's break it down. To find out how much NaCl can be produced, we need to consider the balanced chemical equation for the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
Looking at the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water.
To determine how much NaCl can be produced, we need to find the number of moles of NaOH we have. To do this, we use the molar mass of NaOH:
Molar mass of NaOH = 22.99 g/mol (sodium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 39.99 g/mol
Now we can calculate the number of moles of NaOH:
Number of moles of NaOH = mass of NaOH / molar mass of NaOH
= 3.21 g / 39.99 g/mol
≈ 0.0803 mol NaOH
Since the reaction is 1:1 between NaOH and NaCl, the number of moles of NaCl produced will be the same as the number of moles of NaOH:
Number of moles of NaCl = 0.0803 mol NaOH
Finally, we can calculate the mass of NaCl produced using the molar mass of NaCl:
Molar mass of NaCl = 22.99 g/mol (sodium) + 35.45 g/mol (chlorine) = 58.44 g/mol
Mass of NaCl = number of moles of NaCl * molar mass of NaCl
= 0.0803 mol * 58.44 g/mol
≈ 4.69 g NaCl
So, if the student adds excess HCl to the 3.21 g of NaOH, approximately 4.69 g of NaCl will be produced.
To determine how much NaCl will be produced in the reaction between HCl and NaOH, we first need to write a balanced chemical equation for the reaction:
HCl + NaOH -> NaCl + H2O
From the chemical equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl.
Step 1: Convert the given mass of NaOH to moles.
The molar mass of NaOH (sodium hydroxide) is:
23.0 g/mol (Na) + 16.0 g/mol (O) + 1.0 g/mol (H) = 40.0 g/mol
Using the given mass of NaOH (3.21 g) and its molar mass, we can calculate the number of moles of NaOH:
moles of NaOH = mass of NaOH / molar mass of NaOH
= 3.21 g / 40.0 g/mol
≈ 0.0802 mol
Step 2: Determine the limiting reactant.
Since the student adds an excess of HCl, the limiting reactant will be NaOH, and all of it will be consumed in the reaction.
Step 3: Calculate the moles of NaCl produced.
Since the reaction is 1:1 between NaOH and NaCl, the moles of NaCl produced will be the same as the moles of NaOH used.
moles of NaCl = moles of NaOH = 0.0802 mol
Step 4: Convert moles of NaCl to mass.
The molar mass of NaCl (sodium chloride) is:
23.0 g/mol (Na) + 35.5 g/mol (Cl) = 58.5 g/mol
Using the moles of NaCl (0.0802 mol) and its molar mass, we can calculate the mass of NaCl produced:
mass of NaCl = moles of NaCl × molar mass of NaCl
= 0.0802 mol × 58.5 g/mol
≈ 4.69 g
Therefore, approximately 4.69 grams of NaCl will be produced in the reaction between HCl and NaOH when 3.21 grams of NaOH are used.
To determine how much NaCl can be produced in the reaction between HCl and NaOH, we need to use stoichiometry. The balanced chemical equation for the reaction is:
HCl + NaOH → NaCl + H2O
We can see from the equation that the stoichiometric ratio between HCl and NaCl is 1:1. This means that for every 1 mole of HCl reacted, we produce 1 mole of NaCl.
To calculate the moles of NaOH, we need to convert the given mass (3.21 g) into moles. We can do this using the molar mass of NaOH, which is the sum of the atomic masses of Na (22.99 g/mol), O (16.00 g/mol), and H (1.01 g/mol):
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Now, we can calculate the number of moles of NaOH:
Moles of NaOH = mass of NaOH / molar mass of NaOH
= 3.21 g / 40.00 g/mol
= 0.08025 mol
Since there is an excess of HCl, all the NaOH will be reacted. Therefore, the moles of NaCl produced will be equal to the moles of NaOH reacted.
Moles of NaCl produced = Moles of NaOH reacted = 0.08025 mol
Finally, to calculate the mass of NaCl produced, we can use the molar mass of NaCl, which is 58.44 g/mol:
Mass of NaCl produced = Moles of NaCl produced × molar mass of NaCl
= 0.08025 mol × 58.44 g/mol
= 4.69 g
Therefore, if the student adds excess HCl to the 3.21 g of NaOH, approximately 4.69 g of NaCl will be produced.