Asolution is 40.0% by volume benzene (c6h6) in carbon tetrachloride at 20c. the vapor pressure of pure benzene at this temperature is 74.61 mmhg and its density is 0.87865 g/cm3; the vapor pressure of pure carbon tetrachloride is 91.32 mmhg and its density is 1.5940 g/cm3. if this solution is ideal, its total vapor pressure at 20c is

84.30 mm Hg

Explanation:

In 100 cm³ of solution we have: 40 cm³ C6H6 and 60 cm³ CCl4. Given the densities we can calculate their masses and number of moles, and since by Raoult´s law

Ptotal = XAPºA + XBPºB

where XA= mol fraction =na/(na +nb) and PºA vapor pressure pure of pure component A

m C6H6 = 40 cm³ x 0.87865 g/cm³ = 35.146 g

mol C6H6 = 35.146 g/ 78.11 g/mol = 0.45 mol

mass CCl4 = 60 cm³ x 1.5940 g/cm³ = 95.640 g

mol CCl4 = 95.640 g / 153.82 g/mol = 0.62 mol

mol tot = 1.07

XC6H6 = 0.45/ 1.07 = 0.42      XCCl4 = 0.62/1.07 =0.58

Ptot (mmHg) = 0.42 x 74.61 + .58 x 91.32 = 84.30 mmHg

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2