How many grams of NaCl will be produced when 22.85 g of HCl are neutralized by an excess of NaOH according to the equation below?
HCl + NaOH �¨ H2O + NaCl
ese culo bota caca
To determine the number of grams of NaCl produced when 22.85 g of HCl is neutralized, we need to use the balanced chemical equation and the molar masses of the compounds involved.
The balanced equation is:
HCl + NaOH ⟶ H2O + NaCl
From the equation, we can see that one mole of HCl reacts with one mole of NaCl. So, we need to convert the grams of HCl to moles using its molar mass.
The molar mass of HCl is 36.46 g/mol. Therefore, dividing the given mass by the molar mass gives us:
22.85 g HCl / 36.46 g/mol = 0.6272 mol HCl
Since the reaction is stated to have an excess of NaOH, the same number of moles of NaCl will be produced.
To convert the moles of NaCl to grams, we need to multiply the number of moles by the molar mass of NaCl. The molar mass of NaCl is 58.44 g/mol.
0.6272 mol NaCl × 58.44 g/mol = 36.59 g NaCl
Therefore, approximately 36.59 grams of NaCl will be produced when 22.85 grams of HCl are neutralized.
To determine the number of grams of NaCl produced, we need to use stoichiometry, which is the numerical relationship between the coefficients in a balanced chemical equation.
First, we should look at the balanced equation: HCl + NaOH → H2O + NaCl
The balanced equation tells us that one mole of HCl reacts with one mole of NaOH to form one mole of NaCl.
Now, let's calculate the moles of HCl we have:
Given mass of HCl = 22.85 g
Molar mass of HCl = 36.46 g/mol (from periodic table)
Moles of HCl = mass of HCl / molar mass of HCl
Moles of HCl = 22.85 g / 36.46 g/mol
Moles of HCl ≈ 0.627 mol
Since the equation tells us that 1 mole of HCl reacts to form 1 mole of NaCl, we can conclude that 0.627 moles of HCl will produce 0.627 moles of NaCl.
Now, let's calculate the mass of NaCl produced:
Molar mass of NaCl = 58.44 g/mol (from periodic table)
Mass of NaCl = moles of NaCl × molar mass of NaCl
Mass of NaCl = 0.627 mol × 58.44 g/mol
Mass of NaCl ≈ 36.58 g
Therefore, when 22.85 g of HCl are neutralized by an excess of NaOH, approximately 36.58 grams of NaCl will be produced.