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?
A) 4.00 g
B) 2.49 g
C) 146 g
D) 284 g
E) None of the above.
Please help!!! I think the the answer is C but I am not quite sure.
You are correct but do you know why? Post how you decided the answer was 146 if you want me to critique your work.
Well, let's do some calculations to determine the answer.
First, we need to determine how many moles of H2SO4 we have. We're given that the mass of H2SO4 is 393 g and we can find the molar mass of H2SO4 to be 98.09 g/mol.
Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4
Moles of H2SO4 = 393 g / 98.09 g/mol ≈ 4 moles
Next, we need to determine how many moles of HCl can be produced from 4 moles of H2SO4. From the balanced chemical equation, we can see that the ratio of H2SO4 to HCl is 1:2.
Moles of HCl = 2 × Moles of H2SO4
Moles of HCl = 2 × 4 moles = 8 moles
Finally, we need to determine the mass of HCl produced from the moles of HCl.
Mass of HCl = Moles of HCl × Molar mass of HCl
Mass of HCl = 8 moles × 36.46 g/mol ≈ 288 g
Hmm, it seems like my calculations don't match the answer choices provided. Maybe I made a mistake. Let me double-check my calculations.
To determine the amount of HCl that can be prepared, we need to use mole-to-mole ratios and the molar masses of the substances involved.
First, let's calculate the number of moles of H2SO4:
Molar mass of H2SO4 = (2 x 1.008 g/mol) + (32.06 g/mol) + (4 x 16.00 g/mol)
= 2.016 g/mol + 32.06 g/mol + 64.00 g/mol
= 98.076 g/mol
Given mass of H2SO4 = 393 g
Number of moles of H2SO4 = mass / molar mass
= 393 g / 98.076 g/mol
≈ 4.00 moles (rounded to two decimal places)
Next, let's calculate the number of moles of NaCl:
Given moles of NaCl = 4.00 moles
According to the balanced chemical equation, the mole ratio between H2SO4 and HCl is 1:2. This means that for every 1 mole of H2SO4, we can produce 2 moles of HCl.
Using this information, we can calculate the moles of HCl produced:
Moles of HCl = moles of NaCl x (moles of HCl / moles of NaCl)
= 4.00 moles x (2 moles HCl / 1 mole NaCl)
= 8.00 moles
Finally, let's calculate the mass of HCl:
Molar mass of HCl = 1.008 g/mol + 35.45 g/mol
= 36.46 g/mol
Mass of HCl = moles of HCl x molar mass of HCl
= 8.00 moles x 36.46 g/mol
= 291.68 g
Therefore, the correct answer is not provided in the options (E), as it should be 291.68 g of HCl.
To determine the number of grams of HCl that can be prepared, we first need to find the limiting reactant.
1. Start by calculating the number of moles of NaCl using the given mass of 4.00 moles:
Given mass of NaCl = 4.00 moles
2. Next, find the number of moles of H2SO4 using the given mass of 393 g and its molar mass:
Molar mass of H2SO4 = 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol
Moles of H2SO4 = 393 g / 98.09 g/mol
3. Now, we can determine the stoichiometric ratio of NaCl to HCl from the balanced chemical equation:
2 NaCl(s) + H2SO4(l) ------> 2 HCl(g) + Na2SO4(s)
The ratio is 2:2, meaning that for every 2 moles of NaCl, 2 moles of HCl are produced.
4. Calculate the maximum moles of HCl that can be produced from the limiting reactant (the reactant with fewer moles):
Moles of HCl = 4.00 moles NaCl * (2 moles HCl / 2 moles NaCl) = 4.00 moles HCl
5. Finally, convert the moles of HCl to grams using its molar mass:
Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Mass of HCl = 4.00 moles HCl * 36.46 g/mol = 145.84 g
Therefore, the correct answer is C) 146 g.
Well done on correctly selecting the answer!