Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) What minimum volume of 0.400 M potassium iodide solution is required to completely precipitate all of the lead in 310.0 mL of a 0.112 M lead(II) nitrate solution?

0.1736 L or 173.6 ml

Explanation:

Number of moles of lead II nitrate is obtained by;

Number of moles = concentration × volume of solution

Concentration= 0.112 M

Volume of solution= 310 ml

n= 0.112 × 310/1000

n= 0.03472 moles

From the reaction equation;

2 moles of potassium iodide reacted with 1 mole of lead II nitrate

x moles of potassium iodide will react with 0.03472 moles of lead II nitrate

x= 2 × 0.03472 moles= 0.06944 moles of potassium iodide

Volume of potassium iodide solution = number of moles/ concentration = 0.06944/ 0.4

Volume of potassium iodide solution= 0.1736 L or 173.6 ml

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.