1) In an experiment, 7.36 g of zine was heated with 6.45 g of sulfur. Assume tha
ID: 559114 • Letter: 1
Question
1) In an experiment, 7.36 g of zine was heated with 6.45 g of sulfur. Assume that t according to the equation 8Zn + Ss-NZns. what amount of zanc sulfide was produced? How grams of the excess reactant remain after the reaction is completed. Write the balanced molecular equation and the net ionic equation for each of the following Ba(OH)2 (aq) + HSO, (a) many 2) reactions. a. c. LiCl (aq) + AgNO3 (aq) 3) For each of the following reaction, write the balance molecular equation and then the net ionic equation. a. Aqueous solution of aluminum sulfate and sodium hydroxide are mixed. b. Calcium carbonate and nitric acid are mixe c. Nitrous acid and lithium hydroxide are mixed. 4) The fuel hydrazine can be produced by the reaction of solution of sodium hypochlorite ammonia. NaCIO (aq) +2NH: (aq) N2H4(aq) + NaCl (aq) +H,0 (1) f 750.0 mL of 0.806 M NaC10 is mixed with excess ammonia , how many moles of hydrazine can be formed? If final volume of the resulting solution is l 25What will be the molarity of hydrazine 5) Many common titrations involve the reaction of acid and base. If 24.74 ml. of 0.503 MNaOH solution is used to titrate a 15.00 mL sample of sulfuric acid, H:SO, what is the sulfuric acid in the solution? Whahis h Meleyo in grams, of 2NaOH (agwtso. (aq) NaoSO. (aq) + 2H2O (1) 1+ OHExplanation / Answer
Answer: (1)
=> The balanced chemical reaction for this synthesis reaction
8Zn(s)+S8(s)8 ZnS(s)
=> The mole ratio between zinc and sulfur is 8:1. This tells you that will always consume eight moles of zinc for every one mole of sulfur that takes part in the reaction.
=> Moreover, you have a 1:1 mole ratio between zinc and zinc sulfide, the product of the reaction. This tells you that the reaction will produce as many moles of zinc sulfide as you have moles of zinc that take part in the reaction.
=> Now find the moles of Zn and S
For zinc, Molar mass=65.38 g/mol and given mass=7.36 g, number of moles=mass/molar mass
=7.36g/65.38g/mol=0.11257 moles Zn
For sulfur (S8), molar mass=258.52 g/mol and given mass=6.45 g, number of moles=mass/molar mass
=6.45g/256.52g/mol=0.025144 moles S8
=> Now, you need to use the mole ratio that exists between the two reactants to determine whether or not you have enough moles of zinc to react with that much moles of sulfur.
0.025144moles S8(8 moles Zn/1mole S8)=0.20115 moles Zn
=> Since you have fewer moles of zinc than the complete consumption of the sulfur would have required, you can say that zinc will act a a limiting reagent (Since we have only 0.11257 moles of Zn)
=> More specifically, only 0.11257moles Zn(1 mole S8/8moles Zn)=0.014071 moles S8
will actually take part in the reaction, the rest will be in excess.
=>Now, if zinc is completely consumed by the reaction, it follows that you will end up with
0.11257moles Zn(8 moles ZnS/8moles Zn)=0.11257 moles ZnS
Finally, use zinc sulfide's molar mass (97.474 g/mol) to determine how many grams would contain this many moles
=0.11257moles ZnSx97.474 g1mole ZnS=10.97 g.
Therefore, Zinc is the limiting reagent. The reaction produces 10.97 g of zinc sulfide.
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