Areaction is performed in a vessel attached to a closed-tube manometer. before the reaction, the levels of mercury in the two sides of the manometer were at the same height. as the reaction proceeds, a gas is produced. at the end of the reaction, the height of the mercury column on the vacuum side of the manometer has risen 43.96 cm and the height on the side of the manometer connected to the flask has fallen by the same amount. what is the pressure in the apparatus at the end of the reaction?

879.2 mmHg = 879.2 Torr = 1.16 atm

Explanation:

A mercury manometer measures the pressure by the difference of the height in the two mercury columns. In this case, if one column had decreased, and the other one increased, the high difference will be 2x43.96 cm = 87.92 cm = 879.2 mm.

Thus, the pressure is equal to 879.2 mmHg, or 879.2 Torr.

1 atm 760 Torr

x 879.2 Torr

By a simple direct three rule

760x = 879.2

x = 1.16 atm

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