Question

­­2K + 2HBr → 2 KBr + H2

When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)

●a). What is the limiting reactant?
●b.)What is the excess reactant?
­­2K + 2HBr → 2 KBr + H2

When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)

●a). What is the limiting reactant?
●b.)What is the excess reactant?


­­2K + 2HBr → 2 KBr + H2

When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)

●a). What is the limiting reactant?
●b.)What is the excess reactant?
●C.)How much product is produced?

Answer 1

`\large \boxed{\text{a) HBr; b) K; c) 0.0503 g}}`

Step-by-step explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 39.10 80.41 2.016

2K + 2HBr ⟶ 2KBr + H₂

m/g: 5.5 4.04

a) Limiting reactant

(i) Calculate the moles of each reactant

`\text{Moles of K} = \text{5.5 g} * \frac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} * \frac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}`

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:

The molar ratio of H₂:K is 1:2.

From HBr:

The molar ratio of H₂:HBr is 3:2.

`\text{Moles of H}_(2) = \text{0.049 93 mol HBr } * \frac{\text{1 mol H}_(2)}{\text{2 mol HBr}} = \text{0.024 97 mol H}_(2)`

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

`\text{Mass of H}_(2) = \text{0.024 97 mol H}_(2) * \frac{\text{2.016 g H}_(2)}{\text{1 mol H}_(2)} = \textbf{0.0503 g H}_(2)\\\text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$}`