Question

85.0g of barium metal are added to 275ml of 3.55M solution of hydrochloric acid, how many milliliters of hydrogen gas are collected at 18.5oC and 755.5mmHg in this single replacement reaction?

Answer 1

Answer: 0.024 ml of hydrogen gas are collected at
`18.5^0C` and 755.5mmHg in this single replacement reaction

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Ba=(85.0g)/(137g/mol)=0.620moles`

To calculate the number of moles for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution in ml}}`

`3.55M=\frac{\text{Moles of} HCl* 1000}{275ml}\\\\\text{Moles of }HCl=(3.55* 275)/(1000)=0.976mol`

The balanced chemical reaction is:

`Ba+2HCl\rightarrow BaCl_2+H_2`

According to stoichiometry :

2 moles of
`HCl` require = 1 mole of
`Ba`

Thus 0.976 moles of
`HCl` will require=
`(1)/(2)* 0.976=0.488moles` of
`Ba`

Thus
`HCl` is the limiting reagent as it limits the formation of product and
`Ba` is the excess reagent.

As 2 moles of
`HCl` give = 2 moles of
`H_2`

Thus 0.976 moles of
`HCl` give =
`(2)/(2)* 0.976=0.976moles` of
`H_2`

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 755.5 mmHg = 0.994 atm (760 mm Hg = 1 atm )

V = Volume of gas in L = ?

n = number of moles = 0.976

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`18.5^0C=(18.5+273)K=291.5K`

`V=(nRT)/(P)`

`V=(0.994atm* 0.0820 L atm/K mol* 291.5K)/(0.994atm)=24.0L=0.024ml` (1L=1000ml)

Thus 0.024 ml of hydrogen gas are collected at
`18.5^0C` and 755.5mmHg in this single replacement reaction