Question

1. A 50.0 ml-quantity of 1.30 M hydrobromic acid is poured into 80.0-ml oromic acid is poured into 80.0-mL of 0.350 M calcium sulfide. The gaseous product is collected over water at at 680.0 mm He and 30.-C. All other ions remain dissolved. The vapor press 31.8 mm Hg. To receive full credit, use R=0.0821 L'atm.mol- a. Write the balanced equation for the reaction. Include states equation for the reaction. Include states of matter. No points are awarded; this is to help you [O pts correctly solve the problem. dissolved. The vapor pressure of water at 30.0*C i: 2HBrlag) + Cas (aug) - CaBrelag) + H₂J ig) 0. Determine the maximum quantity of moles of gaseous product produced that could be produced. Circle your answer (show all work using dimensional analysis, discrete calculations will not earn credit; put lines under the last allowed digits.) [8 p c. What volume, in liters, of gas is produced? (Correct number of significant figures in answer here.)

Answer 1

a)

Balanced reaction

2 HBr (aq) + CaS (aq)  $\to$ H₂S (g) + CaBr₂ (aq)

b)

Mmoles of HBr = volume× molarity = 50.0×1.30 = 65

Mmoles of CaS = 80×0.35 = 28

(moles of CaS /moles of HBr ) = $\frac{1}{2}$

65 mmoles of HBr reacts with = 65× _{$\frac{1}{2}$} = 32.5 mmoles CaS.

Hence, limiting reagent is CaS.

Mole ratio of CaS and gaseous product (H₂S) = 1:1

Hence, mmoles of H₂S formed = mmoles of CaS = 28.

Or, moles of H₂S can be formed = 28×10^-3 = 0.028

C..

Temperature (T ) = 30+273= 303 K

Pressure of gas (P) = Total pressure - vapor pressure of water

=( 680 - 31.8) mm Hg

= 648.2 mm Hg

= 648.2 mm Hg × _{$\frac{1\: atm}{760 \: mm \: Hg}$} = 0.852 atm.

Ideal gas equation is

PV = nRT

Or, V = (nRT/P) = (0.028×0.0821×303/0.852) = 0.82 L.

Hence, volume of the gas is 0.82 L ( 2 significant figure).