Answer:
A) 122 atm
Explanation:
PV = nRT
Solve for P --> P = nRT/V
n = 10.0 mol + 5.0 mol = 15.0 mol
R = 0.08206 L atm / mol K
T = 25 + 273 = 298 K
V = 3.0
P = (15.0)(0.08206)(298) / (3.0) = 122 atm
4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
D= mass/volume so it's 100/68 which equals 1.47cm3
