Question

32) The reaction of one mole of nitrogen gas with three moles of hydrogen gas releases 92 kJ of thermal energy to the surroundings. Which of the following is TRUE? A) This reaction has a negative enthalpy of reaction. B) Complete reaction of two moles of nitrogen gas would release 184 kJ of thermal energy in this reaction. C) This reaction is exothermic. D) All of the above are true. E) None of the above are true. 33) How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of NaS and 12.1 g CuSO? Reaction: Na2S+CuSO 4-Na2SO4+ CuS A) 0.05 B) 5.92 C) 15.45 D) 9.58 E) not enough information 34) What is the complete ionic equation for the reaction of hydrochloric acid with potassium hydroxide? A) H+OHH20 B) H+CI+K+OH- 2H + 02-+K*+Cl C) 2H 2C1+K2+.20H- H20+ K2+2C D) H*+Cl+K+OH H20 + K*+Cl E) none of the above 35) What are the coefficients for the following reaction when it is properly balanced? NaNO 3 *_Ba3 (PO4)2 Na3 PO4Ba(NO3)2 A) 2,3, 6, 1 B) 2, 1, 1,3 C) 6, 1, 3, 2 D) 2, 3, 1, 6 E) none of the above

Answer 1

32)

Since energy is being released in reaction, this is exothermic reaction.

Delta H is negative for exothermic reaction.

For 1 mol, heat evolved = 92 KJ

So, for 2 mol it would be 2*92 KJ = 184 KJ

Answer: D

33)

Molar mass of Na2S,

MM = 2*MM(Na) + 1*MM(S)

= 2*22.99 + 1*32.07

= 78.05 g/mol

mass(Na2S)= 15.5 g

use:

number of mol of Na2S,

n = mass of Na2S/molar mass of Na2S

=(15.5 g)/(78.05 g/mol)

= 0.1986 mol

Molar mass of CuSO4,

MM = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass(CuSO4)= 12.1 g

use:

number of mol of CuSO4,

n = mass of CuSO4/molar mass of CuSO4

=(12.1 g)/(1.596*10^2 g/mol)

= 7.581*10^-2 mol

Balanced chemical equation is:

Na2S + CuSO4 ---> Na2SO4 + CuS

1 mol of Na2S reacts with 1 mol of CuSO4

for 0.1986 mol of Na2S, 0.1986 mol of CuSO4 is required

But we have 7.581*10^-2 mol of CuSO4

so, CuSO4 is limiting reagent

According to balanced equation

mol of Na2S reacted = (1/1)* moles of CuSO4

= (1/1)*7.581*10^-2

= 7.581*10^-2 mol

mol of Na2S remaining = mol initially present - mol reacted

mol of Na2S remaining = 0.1986 - 7.581*10^-2

mol of Na2S remaining = 0.1228 mol

Molar mass of Na2S,

MM = 2*MM(Na) + 1*MM(S)

= 2*22.99 + 1*32.07

= 78.05 g/mol

use:

mass of Na2S,

m = number of mol * molar mass

= 0.1228 mol * 78.05 g/mol

= 9.583 g

Answer: 9.58 g

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