Hydrochloric acid can be prepared using the reaction described by the chemical equation: 2 NaCl(s) + H2SO4(l) ----> 2 HCl(g) + Na2 SO4(s). How many grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl?
This is a limiting reagent problem.
Convert 393 grams H2SO4 to moles H2SO4.
a. Using the coefficients in the balanced equation, convert moles H2SO4 to moles HCl.
b. Using the coefficients in the balanced equation, convert 4.00 moles NaCl to moles HCl.
c. Both numbers can't be right. The correct one to choose is ALWAYS the smaller one.
Convert the smaller number of moles HCl to grams. g = moles x molar mass.
To determine the number of grams of HCl that can be prepared, we need to first find the limiting reactant in the given reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed.
First, let's calculate the molar mass of H2SO4:
H2SO4 = (2 * 1.00784 g/mol) + (32.06 g/mol) + (4 * 16.00 g/mol)
= 98.09 g/mol
Next, we need to calculate the number of moles of H2SO4 using its given mass:
Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4
= 393 g / 98.09 g/mol
≈ 4.00 moles of H2SO4
Now, let's calculate the number of moles of HCl that can be formed from the given moles of NaCl:
From the balanced chemical equation, we can see that the stoichiometry between NaCl and HCl is 2:2. That means for every 2 moles of NaCl, we can obtain 2 moles of HCl.
Since the given moles of NaCl is 4.00 moles, we can assume that we can obtain 4.00 moles of HCl as well.
Finally, let's calculate the mass of HCl using its molar mass:
Molar mass of HCl = 1.00784 g/mol + 35.453 g/mol
= 36.461 g/mol
Mass of HCl = Number of moles of HCl * Molar mass of HCl
= 4.00 moles * 36.461 g/mol
≈ 145.844 g
Therefore, approximately 145.844 grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl.