If 100.0 mL of 0.1010.101 M Na2SO4Na2SO4 is added to 100.0 mL of 0.7790.779 M Pb(NO3)2Pb(NO3)2, how many grams of PbSO4PbSO4 can be produced? Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s) STRATEGY Calculate the initial moles present and add them to a table of initial, change, and end of reaction moles. Determine the limiting reactant using the comparison-of-moles method. Use the limiting reactant to determine the change in moles for each reactant and product. Then, complete the table of moles. Convert the moles of PbSO4PbSO4 produced to the mass of PbSO4PbSO4, in grams. Step 1: Calculate the initial moles present and add them to the table. 0.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO40.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO4 0.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)20.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)2 moles Na2SO4(aq)Na2SO4(aq) ++ Pb(NO3)2(aq)Pb(NO3)2(aq) ⟶⟶ 2NaNO3(aq)2NaNO3(aq) ++ PbSO4(s)PbSO4(s) initial 0.01010.0101 0.07790.0779 0 0 change ? ? ? ? equilibrium ? ? ? ? Step 2: Which reactant is limiting? Pb(NO3)2Pb(NO3)2 Na2SO4
If 100.0 mL of 0.1010.101 M Na2SO4Na2SO4 is added to 100.0 mL of 0.7790.779 M Pb(NO3)2Pb(NO3)2, how many grams of PbSO4PbSO4 can be produced? Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s) STRATEGY Calculate the initial moles present and add them to a table of initial, change, and end of reaction moles. Determine the limiting reactant using the comparison-of-moles method. Use the limiting reactant to determine the change in moles for each reactant and product. Then, complete the table of moles. Convert the moles of PbSO4PbSO4 produced to the mass of PbSO4PbSO4, in grams. Step 1: Calculate the initial moles present and add them to the table. 0.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO40.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO4 0.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)20.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)2 moles Na2SO4(aq)Na2SO4(aq) ++ Pb(NO3)2(aq)Pb(NO3)2(aq) ⟶⟶ 2NaNO3(aq)2NaNO3(aq) ++ PbSO4(s)PbSO4(s) initial 0.01010.0101 0.07790.0779 0 0 change ? ? ? ? equilibrium ? ? ? ? Step 2: Which reactant is limiting? Pb(NO3)2Pb(NO3)2 Na2SO4
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
If 100.0 mL of 0.1010.101 M Na2SO4Na2SO4 is added to 100.0 mL of 0.7790.779 M Pb(NO3)2Pb(NO3)2, how many grams of PbSO4PbSO4 can be produced?
Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)
STRATEGY
-
Calculate the initial moles present and add them to a table of initial, change, and end of reaction moles.
-
Determine the limiting reactant using the comparison-of-moles method.
-
Use the limiting reactant to determine the change in moles for each reactant and product. Then, complete the table of moles.
-
Convert the moles of PbSO4PbSO4 produced to the mass of PbSO4PbSO4, in grams.
Step 1: Calculate the initial moles present and add them to the table.
0.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO40.101 mol Na2SO4L×1 L1000 mL× 100.0 mL=0.0101 mol Na2SO4
0.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)20.779 molPb(NO3)2L×1 L1000 mL× 100.0 mL=0.0779 molPb(NO3)2
| moles | Na2SO4(aq)Na2SO4(aq) | ++ | Pb(NO3)2(aq)Pb(NO3)2(aq) | ⟶⟶ | 2NaNO3(aq)2NaNO3(aq) | ++ | PbSO4(s)PbSO4(s) |
|---|---|---|---|---|---|---|---|
| initial | 0.01010.0101 | 0.07790.0779 | 0 | 0 | |||
| change | ? | ? | ? | ? | |||
| equilibrium | ? | ? | ? | ? |
Step 2: Which reactant is limiting?
Pb(NO3)2Pb(NO3)2
Na2SO4
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