Hydrogen chloride gas can be prepared by the following reaction:
2NaCl(s) + H2SO4(aq) --> 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
To determine the number of grams of HCl that can be prepared, we need to use the given number of moles of H2SO4 and NaCl and convert them to the number of moles of HCl, and then to grams using the molar mass of HCl.
First, let's find the number of moles of HCl that can be prepared from 2.00 mol H2SO4. By examining the balanced chemical equation, we can see that the stoichiometric ratio between H2SO4 and HCl is 1:2.
So, if we have 2.00 mol of H2SO4, we can expect to get twice as many moles of HCl.
2.00 mol H2SO4 x 2 mol HCl / 1 mol H2SO4 = 4.00 mol HCl
Next, let's find the number of moles of HCl that can be prepared from 2.56 mol NaCl. Again, by examining the balanced chemical equation, we can see that the stoichiometric ratio between NaCl and HCl is 2:2.
So, if we have 2.56 mol of NaCl, we can expect to get the same number of moles of HCl.
2.56 mol NaCl x 2 mol HCl / 2 mol NaCl = 2.56 mol HCl
Now, let's add up the moles of HCl obtained from both H2SO4 and NaCl:
4.00 mol HCl (from H2SO4) + 2.56 mol HCl (from NaCl) = 6.56 mol HCl
Finally, let's convert the moles of HCl to grams using the molar mass of HCl. The molar mass of HCl is approximately 36.46 g/mol.
6.56 mol HCl x 36.46 g/mol = 239.5776 g HCl
Therefore, approximately 239.58 grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl.
To find the grams of HCl that can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl, we need to determine the limiting reactant first.
1. For H2SO4:
Using the balanced equation, we see that 1 mole of H2SO4 produces 2 moles of HCl. Therefore, 2.00 mol of H2SO4 will produce:
2.00 mol H2SO4 x (2 mol HCl / 1 mol H2SO4) = 4.00 mol HCl
2. For NaCl:
Using the balanced equation, we see that 2 moles of NaCl produce 2 moles of HCl. Therefore, 2.56 mol of NaCl will produce:
2.56 mol NaCl x (2 mol HCl / 2 mol NaCl) = 2.56 mol HCl
Since the stoichiometric ratio for NaCl is equal to the stoichiometric ratio for HCl, we can conclude that NaCl is the limiting reactant.
3. To find the mass of HCl produced from 2.56 mol HCl, we need to know the molar mass of HCl. The molar mass of HCl is 36.461 g/mol.
Mass of HCl = Moles of HCl x Molar mass of HCl
Mass of HCl = 2.56 mol x 36.461 g/mol = 93.504 g
Therefore, you can prepare 93.504 grams of HCl from 2.00 mol H2SO4 and 2.56 mol NaCl.
I have rewritten the equation without phases to save space. This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants.
2NaCl + H2SO4 --> 2HCl + Na2SO4
Usint 2.00 mols H2SO4 and all of the NaCl needed will give you 2.00 x (2 mols HCl/1 mol H2SO4) = 2.00 x 2 = 4.00 mol HCl
Using 2.56 mols NaCl and all of the H2SO4 needed will give you
2.56 x (2 mols HCl/2 mols NaCl) = 2.56 mols HCl.
Note the values are different (they usually are in LR problems) so one must be incorrect; the correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR. SO you will obtain 2.56 mol HCl.
Convert that value to grams. g = mols x molar mass = ?