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Application - Equations and reacting masses
1) Balance the equation: KCI + F2 → KF + Cl2
2) How many moles of F2 are required to produce 11.8 moles of KF?
3) What is the mass of 11.8 moles of KF?
4) What is the mass of 5 moles of F2?
5) 10g of F2 are used. What mass of KCl is required?
6) What mass of KF will be produced?
7) What mass of Cl2 will be produced?
8) A student wants to make 39.2g of KF. What mass of KCl and F2 are required?

Respuesta :

cristisp93

1) 2 KCl + F₂ → 2 KF + Cl₂

2) 5.9 moles of F₂

3) 684.4 g of KF

4) 190 g of F₂

5) 0.526 g of KCl

6) 30.508 g of KF

7) 18.673 g of Cl₂

8) 50.362 g of KCl and 12.844 g of F₂

Explanation:

1) To balance the equation the number and type of moles entering the reaction have to be equal to the number and type of moles leaving the reaction.

2 KCl + F₂ → 2 KF + Cl₂

2) Knowing the chemical reaction we devise the following reasoning:

if        1 mole of F₂ produces 2 moles of KF

then   X moles of F₂ produces 11.8 moles of KF

X = (1 × 11.8) / 2 = 5.9 moles of F₂

3) mass = number of moles × molecular weight

mass of KF = 11.8 × 58 = 684.4 g

4) mass = number of moles × molecular weight

mass of F₂ = 5 × 38 = 190 g

5)  number of moles = mass / molecular weight

number of moles of  F₂ = 10 / 38 = 0.263 moles

Knowing the chemical reaction we devise the following reasoning:

if        1 mole of F₂ react with 2 moles of KCl

then   0.263 moles of F₂ produces Y moles of KCl

Y = (0.263 × 2) / 1 = 0.526 g of KCl

6) Knowing the chemical reaction we devise the following reasoning:

if        1 mole of F₂ produces 2 moles of KF

then   0.263 moles of F₂ produces Z moles of KF

Z = (0.263 × 2) / 1 = 0.526 moles of KF

mass = number of moles × molecular weight

mass of KF = 0.526 × 58 = 30.508 g

7) Knowing the chemical reaction we devise the following reasoning:

if        1 mole of F₂ produces 1 mole of Cl₂

then   0.263 moles of F₂ produces W moles of Cl₂

W = (0.263 × 1) / 1 = 0.263 moles of Cl₂

mass = number of moles × molecular weight

mass of Cl₂ = 0.263 × 71 = 18.673 g

8) number of moles = mass / molecular weight

number of moles of KF = 39.2 / 58 = 0.676 moles

Knowing the chemical reaction we devise the following reasoning:

if        2 moles of KF are produced from 2 moles of KCl and 1 mole of F₂

then  0.676 moles of KF are produced from A moles of KF and B moles of F₂

A = (0.676 × 2) / 2 = 0.676 moles of KCl

B = (0.676 × 1) / 2 = 0.338 moles of F₂

mass = number of moles × molecular weight

mass of KCl = 0.676 × 74.5 = 50.362 g

mass of F₂ = 0.338 × 38 = 12.844 g

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