Help with any of these:
21. Aluminium sulphate is often used by gardeners to acidify soil and can be produced by the reaction
2Al(OH)3(aq) + 3 H2SO4(aq) à Al2(SO4)3(aq) + 6 H2O(l)
The volume of 0.275 mol/L H2SO4(aq) needed to react completely with 112 mL of 0.355 mol/l Al(OH)3(aq) is
A. 96.4 mL
B. 145 mL
C. 168 mL
D. 217 mL
22. To produce the anti cancer drug cisplatin (Pt (NH3)2Cl2(s)), a technician reacted NH3(aq) with excess K2PtCl4(aq). The reaction follows
K2PtCl4(aq) + 2 NH3(aq) à Pt(NH3)2Cl2(s) + 2 KCl(aq)
The data was recorded by the technician as follows.
Concentration of NH3(aq) used = 0.450 mol/L
Mass of filter paper = 0.725 g
Mass of filter paper and solid precipitate = 9.544 g
The volume of NH3(aq) used by the technician was
A. 32.7 mL
B. 65.3 mL
C. 131 mL
D. 141 mL
23. When calcium reacts with nitrogen, calcium nitride is formed, as shown in the following balanced chemical equation:
3 Ca(s) + N2(g) Ca3N2(s)
If 24.0 g of calcium and 12.0 g of nitrogen are available for this reaction, the limiting reagent is:
A. calcium
B. nitrogen
C. calcium nitride
D. both calcium and nitrogen will be completely consumed
24. When 45 mL of 0.25 mol/L sulfuric acid is mixed with 25 mL of 0.48 mol/L calcium hydroxide, the solution
A. is acidic
B. is neutral
C. is basic
D. not enough information is given
25. Consider the reaction equation 3 Fe(s) + 4 H2O(l) Fe3O4(s) + 4 H2(g). If 3.0 mol Fe reacts with 1.0 mol H2O in ideal conditions, which statement is true?
A. Fe is in excess.
B. H2O is in excess.
C. Both reactants are completely consumed.
D. Not enough information is given to answer the question.
26. A group of students performed a titration analysis. The results are given in the graph below.
The indicator that the students should use to find the endpoint pH is:
A. methyl orange
B. methyl red
C. bromothymol blue
D. phenolphthalein
27. In a titration, a 20 mL sample of NaOH(aq) was neutralized by 14.9 mL of 0.13 mol/L H2SO4(aq). The concentration of the base is
A. 8.7 mol/L
B. 0.35 mol/L
C. 0.19 mol/L
D. 0.048 mol/L
28. For complete neutralization, 15.0 mL of 0.35 mol/L sodium hydroxide solution was required to react with 0.425 g of an acid. The possible identity of the acid is.
A. HBr (aq)
B. HCl(aq)
C. HNO3(aq)
D. H2SO4(aq)
29. A lab group decides to test a manufacturer’s claim that an antacid product contains 200 mg of sodium bicarbonate. Each tablet was placed into an Erlenmeyer flask, 10.00 mL of water was added, and then the tablets were crushed. An indicator was added to each flask, and then each sample was titrated with 0.200 mol/L hydrochloric acid. The titration evidence was collected in the table below
Titration of 10.00 mL of NaHCO3(aq) with 0.200 mol/L HCl(aq)
Trial
1
2
3
4
Final burette reading (mL)
11.6
21.7
31.7
41.8
Initial burette reading (mL)
0.6
11.6
21.7
31.7
Volume of HCl(aq) added (mL)
Indicator colour
red
orange
orange
orange
The calculated mass of sodium bicarbonate in the antacid tablet is:
A. 169 mg
B. 173 mg
C. 84.6 mg
D. 86.5 mg
30. During a titration, results were recorded for three trials for the complete neutralization of 0.100 mol/L potassium carbonate solution with hydrochloric acid
The Volume of HCl(aq) Required to React With a Given Volume of K2CO3(aq)
Volume (mL)
Trial 1
Trial 2
Trial 3
K2CO3(aq)
20.0
20.0
20.0
HCI(aq)
22.3
22.5
22.4
From the data, it can be calculated that the concentration of the acid is
A. 4.46 X 10–1 mol/L
B. 1.79 X 10–1 mol/L
C. 4.46 X 10–2 mol/L
D. 1.79 X 10–2 mol/L
31. When a solution of H2S(aq) is mixed with a solution of Pb(NO3)2(aq), the ionic equation of the above reaction is
a. Pb2+ (aq) + 2NO3–(aq) Pb(NO3)2(aq)
b. S2–(aq) + Pb2+ (aq) PbS(s)
c. 2H+(aq) + S2–(aq) + Pb2+ (aq) + 2NO3–(aq) PbS(s) + 2H+(aq) + 2NO3–(aq)
d. S2(aq) + Pb2+ (aq) + 2NO3(aq) PbS(s) + 2NO3(aq)
32.
In the Haber process, ammonia is formed by the reaction of nitrogen with hydrogen under high pressure and high temperature. The reaction is expressed as follows:
If 2.00 g of N2(g) reacts with excess hydrogen to produce 1.85 g of ammonia, what will the percentage yield be for the reaction?
a. 75.9% c. 77.9%
b. 76.1% d. 82.2%
33. A solution of NaCl(aq) is added dropwise to a 10.0 mL sample of water from an industrial site that contains lead(II) nitrate, Pb(NO3)2(aq). A white precipitate forms immediately. What is the net ionic equation for the reaction that occurred?
A. Na+(aq) + Cl-(aq) ® NaCl(s)
B. Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq) ® PbCl2(s) + 2Na+(aq) + 2Cl-(aq)
C. Pb2+(aq) + 2NO3-(aq) + 2Na+(aq) + 2Cl-(aq) ® Pb2+(aq) + 2Cl-(aq) + 2NaCl(s)
D. Pb2+(aq) + 2Cl-(aq) ® PbCl2(s)
NR1 When 12.71 g of copper react with 8.000 g of sulfur, 14.72 g of copper(I) sulphide is produced. What is the percentage yield in this reaction?
2Cu(s) + S(g) ® Cu2S(s)
NR2. What mass of the reagent in excess will remain after the reaction in which
0.3604 g of H2O(l) react with 0.8000 g of O2(g)? 2H2O(l) + O2(g) ---> 2H2O2
Two suggestions to help.
1. You get help more quickly if you post just one at a time.
2. Most of the problems are either stoichiometric or limiting reagent problems. Here are examples of each; just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
The titration problems can be solved by a three step process.
25 mL A + 50 mL of 0.1 B. What is molarity of A.
1. Convert B to mols. mols = M x L = ?
2. Use the coefficients in the balanced equation to convert mols B to mols A.
3. Then M A = mols A/L A.
b.
Sure, I can help you with these questions. Let's go step by step.
21. To find the volume of H2SO4(aq) needed, we need to use the stoichiometry of the balanced equation. According to the equation:
2Al(OH)3(aq) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
From the balanced equation, we can see that 2 moles of Al(OH)3 react with 3 moles of H2SO4.
Given that the concentration of Al(OH)3 is 0.355 mol/L and the volume is 112 mL, we can calculate the moles of Al(OH)3 as follows:
Moles of Al(OH)3 = concentration * volume
Moles of Al(OH)3 = 0.355 mol/L * 0.112 L (convert mL to L) = 0.03976 mol
Using the mole ratio from the balanced equation, we can determine the moles of H2SO4 needed:
Moles of H2SO4 = (2/2) * (3/2) * moles of Al(OH)3
Moles of H2SO4 = 0.05964 mol
Now, let's find the volume of H2SO4:
Volume of H2SO4 = Moles of H2SO4 / Concentration
Volume of H2SO4 = 0.05964 mol / 0.275 mol/L = 0.2169 L (convert to mL)
Volume of H2SO4 = 216.9 mL
Therefore, the volume of 0.275 mol/L H2SO4(aq) needed to react completely with 112 mL of 0.355 mol/L Al(OH)3(aq) is approximately 217 mL.
Answer: D. 217 mL
22. Similarly, to find the volume of NH3(aq) used, we need to use the stoichiometry of the balanced equation. According to the equation:
K2PtCl4(aq) + 2 NH3(aq) → Pt(NH3)2Cl2(s) + 2 KCl(aq)
From the balanced equation, we can see that 1 mole of K2PtCl4 reacts with 2 moles of NH3.
Given that the concentration of NH3(aq) is 0.450 mol/L and the mass of the filter paper and solid precipitate is 9.544 g, we can calculate the moles of NH3 as follows:
Moles of NH3 = concentration * volume
Moles of NH3 = 0.450 mol/L * volume (in Liters)
To find the volume in Liters, we need to convert the mass of the filter paper and solid precipitate to moles of Pt(NH3)2Cl2:
Moles of Pt(NH3)2Cl2 = (mass of filter paper and solid precipitate - mass of filter paper) / molar mass of Pt(NH3)2Cl2
Given that the molar mass of Pt(NH3)2Cl2 is 300 g/mol:
Moles of Pt(NH3)2Cl2 = (9.544 g - 0.725 g) / 300 g/mol
Moles of Pt(NH3)2Cl2 = 0.02964 mol
Using the mole ratio from the balanced equation, we can determine the moles of NH3 used:
Moles of NH3 = (1/2) * moles of Pt(NH3)2Cl2
Moles of NH3 = 0.02964 mol / 2
Moles of NH3 = 0.01482 mol
Now, let's find the volume of NH3:
Volume of NH3 = Moles of NH3 / Concentration
Volume of NH3 = 0.01482 mol / 0.450 mol/L = 0.03293 L (convert to mL)
Volume of NH3 = 32.93 mL
Therefore, the volume of NH3(aq) used by the technician is approximately 32.7 mL.
Answer: A. 32.7 mL
Please let me know if you need help with the remaining questions.
21. To find the volume of H2SO4(aq) needed to react completely with the given volume and concentration of Al(OH)3(aq), we need to use stoichiometry.
First, convert the given volume of Al(OH)3(aq) to moles:
0.112 L * 0.355 mol/L = 0.03976 mol Al(OH)3
From the balanced equation, the stoichiometric ratio between Al(OH)3 and H2SO4 is 2:3. This means that for every 2 moles of Al(OH)3, we need 3 moles of H2SO4.
Using stoichiometry, we can calculate the moles of H2SO4 required:
0.03976 mol Al(OH)3 * (3 mol H2SO4 / 2 mol Al(OH)3) = 0.05964 mol H2SO4
Finally, calculate the volume of H2SO4(aq) using its concentration:
Volume = moles / concentration
Volume = 0.05964 mol / 0.275 mol/L = 0.2169 L = 216.9 mL
Therefore, the volume of 0.275 mol/L H2SO4(aq) needed to react completely with 112 mL of 0.355 mol/L Al(OH)3(aq) is approximately 216.9 mL. Therefore, the correct answer is D. 217 mL.
22. To find the volume of NH3(aq) used, we need to use stoichiometry.
From the balanced equation, the stoichiometric ratio between K2PtCl4 and NH3 is 1:2. This means that for every 1 mole of K2PtCl4, we need 2 moles of NH3.
The moles of NH3 used can be calculated using stoichiometry:
0.450 mol/L * 0.01000 L = 0.00450 mol NH3
Since NH3 is in excess, all of it will be used in the reaction.
Therefore, the volume of NH3(aq) used by the technician is 0.01000 L or 10.00 mL. Therefore, the correct answer is not given in the options provided.
Let me know if you need explanation for the remaining questions.