Question

How many liters of a 2.25 molar hydrobromic acid (HBr) solution would be needed to react completely with 100.0 grams of calcium metal?

Ca (s) + 2HBr (aq) CaBr2 (aq) + H2 (g)

Answer 1

Answer: 2.22 L

Explanation: To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}` ....(1)

For calcium:

Mass of calcium given = 100 g

Molar mass of calcium = 40 g/mol

Putting values in above equation, we get:

`\text{Moles of calcium}=(100g)/(40g/mol)=2.5mol`

For the given reaction:

`Ca(s)+2HBr(aq)\rightarrow CaBr_2(aq)+H_2(g)`

By Stoichiometry of the reaction,

1 mole of calcium reacts 2 moles of hydrogen bromide

So, 2.5 moles of calcium will react with =
`(2)/(1)* 2.5=5moles` of hydrogen bromide.

Molarity is defined as the number of moles of solute dissolved per liter of solution.

Formula used :

`Molarity=(n)/(V_s)`

where,

n = moles of solute
`HBr` = 5 moles

`V_s` = volume of solution in liter = ?

`2.25=(5moles)/(V_s)`

`V_s=2.22L`

Therefore, the volume of solution in Liters is 2.22.

Answer 2

We can solve by dimensional analysis. We are already given with the concentration of the acid, the amount of the metal, and the balanced chemical reaction, so:
100 g Ca (1 mol Ca / 40 g Ca) (2 mol HBr / 1 mol Ca) (1 L HBr / 2.25 mol HBr)
= 2.22 L HBr
It would require 2.22 L of HBr to react completely 100 g of Ca.