When 1.06 grams of Ag2NO3 (FW=106g/mol) reacts with 3.65 grams of HCl (FW=36.5g/mol), how much H2CO3 (FW= 62g/mol) in grams will be formed? Which reactant is the limiting reactant?
You need to proof the problem because something ain't right.
There is no such thing as Ag2NO3. You react with AgNO3 with HCl and you don't get H2CO3. I'll bet you meant Ag2CO3. So read what you posted. Correct. Repost.
Well, let me calculate it for you while keeping a straight face... Just kidding! Let's dive into the problem.
To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio between them. So, let's calculate the moles of each reactant:
Moles of Ag2NO3 = (1.06g) / (106g/mol) = 0.01 moles
Moles of HCl = (3.65g) / (36.5g/mol) = 0.1 moles
Now, let's use the balanced equation to determine the stoichiometric ratio between Ag2NO3 and H2CO3. From the equation:
2AgNO3 + H2CO3 -> 2HNO3 + Ag2CO3
We can see that 2 moles of Ag2NO3 react with 1 mole of H2CO3. Therefore, the stoichiometric ratio is 2:1.
Now, let's see which reactant is present in a lower amount. Since we have 0.01 moles of Ag2NO3 and 0.1 moles of HCl, it's clear that Ag2NO3 is the limiting reactant.
To find the amount of H2CO3 formed, we need to use the stoichiometric ratio. Since Ag2NO3 and H2CO3 are in a 2:1 ratio, we can say that half the moles of Ag2NO3 will be the moles of H2CO3 formed.
Moles of H2CO3 formed = (0.01 moles of Ag2NO3) / 2 = 0.005 moles
Lastly, we can calculate the grams of H2CO3 formed:
Grams of H2CO3 formed = (0.005 moles) * (62g/mol) = 0.31 grams
Therefore, approximately 0.31 grams of H2CO3 will be formed. And the reactant that limits the reaction is Ag2NO3. Hope that put a smile on your face!
To find the amount of H2CO3 formed and determine the limiting reactant, you need to follow a few steps:
Step 1: Convert the given masses of Ag2NO3 and HCl to moles.
Moles of Ag2NO3 = mass of Ag2NO3 (in grams) / molar mass of Ag2NO3
Moles of HCl = mass of HCl (in grams) / molar mass of HCl
Moles of Ag2NO3 = 1.06 g / 106 g/mol = 0.01 moles of Ag2NO3
Moles of HCl = 3.65 g / 36.5 g/mol = 0.10 moles of HCl
Step 2: Write the balanced chemical equation for the reaction.
2 Ag2NO3 + 2 HCl → 2 AgCl + H2CO3
Step 3: Determine the stoichiometry between Ag2NO3 and H2CO3.
From the balanced equation, we can see that 2 moles of Ag2NO3 react to form 1 mole of H2CO3.
Step 4: Calculate the amount of H2CO3 formed.
Moles of H2CO3 = (moles of Ag2NO3) x (1 mole H2CO3 / 2 moles Ag2NO3)
Moles of H2CO3 = 0.01 moles of Ag2NO3 x (1 mole H2CO3 / 2 moles Ag2NO3) = 0.005 moles of H2CO3
Step 5: Convert moles of H2CO3 to grams.
Mass of H2CO3 = moles of H2CO3 x molar mass of H2CO3
Mass of H2CO3 = 0.005 moles x 62 g/mol = 0.31 grams of H2CO3
Based on the calculations, 0.31 grams of H2CO3 will be formed.
To determine the limiting reactant, compare the moles of Ag2NO3 and HCl. The reactant that produces fewer moles of the desired product (H2CO3 in this case) is the limiting reactant. In this case, 0.01 moles of Ag2NO3 is less than 0.10 moles of HCl, so Ag2NO3 is the limiting reactant.