One way to make chlorine is to let manganese dioxide, MnO2, react with hydrochloric acid according to the following equation.
4HCl + MnO2 Cl2 + MnCl2 + 2H2O

What is the minimum volume (in mL) of 6.91 M HCl needed to prepare 460.9 mL of dry chlorine when the gas is obtained at 21.0 °C and 750 torr?

mL

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter  

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

explanation:

0.07 g/cm³

1.37 g/cm³