Question

78) How many grams of KCl (molar mass = 74.55 g) will be produced from the reaction of 50.0 mL of 0.300 M KOH with excess HCl? KOH aq + HCl aq → KCl aq + H2O(ℓ) A) 1.12 g B) 9.31 g C) 0.876 g D) 6.98 g E) 4.65 g

79) Which of the following compounds has the largest formula mass? A) HCN B) SiH4 C) NO2 D) CCl4 E) BF3

80) Which of the following conditions characterizes a system in a state of chemical equilibrium? A) Product concentrations are greater than reactant concentrations. B) Rate of forward reaction has dropped to zero. C) Concentrations of reactants and products are equal. D) Reactant molecules no longer react with each other. E) Reactants are being consumed at the same rate they are being produced.

81) Elements in group VA of the periodic table ________. A) form ions with a charge of 4+ B) are type II metals C) form ions with a charge of 3- D) form ions with a charge of 2- E) are type I nonmetals

Answer 1

78. The reaction equation is : KOH + HCl ----> KCl + H2O

Moles of KOH = molarity*volume in L = 0.3*0.05 = 0.015 moles

1 mol of KOH give 1 mole of KCl.

Moles of KCl produced = 0.015 mol

Mass of KCl produced = moles*molar mass = 0.015*74.55 = 1.12 g

Answer: a) 1.12 g