For each precipitation reaction, calculate how many grams of the first reactant are necessary to completely react with 17.3 g of the second reactant.
2KI(aq) + Pb(NO3)2(aq)--->PbI2(s) + 2KNO3(aq)
Not looking for the answer I really want to know if I did it correctly
17.3g Pb(NO3)/331.22gPb(NO3) x 2 mol KI x 166g KI = 17.3G KI
Well, it seems like you made a small mistake in your calculation. Let's break it down:
First, you correctly found the molar mass of Pb(NO3)2 to be 331.22 g/mol.
Next, you started setting up the conversion factors, but you switched the reactants in the equation. The coefficient in front of Pb(NO3)2 is 1, not 2, so it should be 1 mol Pb(NO3)2 instead.
Finally, you multiplied by the molar mass of KI as you should, which is 166 g/mol.
So, the corrected calculation would be:
17.3 g Pb(NO3)2 / 331.22 g/mol Pb(NO3)2 x 1 mol KI x 166 g/mol KI = (your answer in grams)
I hope this helps! Keep up the good work, and don't hesitate to ask any more chemistry questions.
To calculate how many grams of the first reactant (KI) are necessary to completely react with 17.3 g of the second reactant (Pb(NO3)2), you need to use the stoichiometry of the reaction.
First, calculate the molar mass of Pb(NO3)2:
Pb(NO3)2 = 207.2 g/mol + (14.01 g/mol + 3(16.00 g/mol))2 = 331.22 g/mol
Now, we can set up the calculation:
(17.3 g Pb(NO3)2) / (331.22 g Pb(NO3)2/mol Pb(NO3)2) × (2 mol KI/1 mol Pb(NO3)2) × (166 g KI/mol KI) = x g KI
Simplifying the calculation:
x = (17.3 g Pb(NO3)2) × (2 mol KI/1 mol Pb(NO3)2) × (166 g KI/331.22 g Pb(NO3)2)
x = 17.3 g Pb(NO3)2 × 2 × 166 g KI / 331.22 g Pb(NO3)2
Now, evaluate the expression:
x = (17.3 × 2 × 166) / 331.22
x = 17.3 × (2 × 166) / 331.22
x = 17.3 × 332 / 331.22
x = 332/19.17
x ≈ 17.288 g KI
So, approximately 17.288 grams of KI are necessary to completely react with 17.3 grams of Pb(NO3)2.
To determine if you calculated correctly, let's go through the steps together:
1. Write down the balanced equation:
2KI(aq) + Pb(NO3)2(aq) ---> PbI2(s) + 2KNO3(aq)
2. Determine the molar mass of Pb(NO3)2:
Pb(NO3)2: 1 Pb atom + 2 N atoms + 6 O atoms = 1 (207.2 g/mol) + 2 (14.0 g/mol) + 6 (16.0 g/mol) = 331.2 g/mol
3. Set up a conversion factor to determine the amount of KI needed:
17.3 g Pb(NO3)2 x (1 mol Pb(NO3)2 / 331.2 g Pb(NO3)2) x (2 mol KI / 1 mol Pb(NO3)2) x (166.0 g KI / 1 mol KI) = ? g KI
Let's solve the calculation:
17.3 g Pb(NO3)2 x (1 mol Pb(NO3)2 / 331.2 g Pb(NO3)2) x (2 mol KI / 1 mol Pb(NO3)2) x (166.0 g KI / 1 mol KI) = 17.3 g KI
Based on the calculation you provided, you have correctly determined that 17.3 g of Pb(NO3)2 reacts with 17.3 g of KI. Well done!