Warm-Up

Use the equation to determine what mass of FeS must react to form 326 g of FeCl2.

iogann1982 [59]

Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

Mass of FeCl2 = 326 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: The balanced equation

FeS + 2 HCl → H2S + FeCl2

Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

4 0

3 years ago

4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.

Bas_tet [7]

Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.

7 0

3 years ago

The illustration depicts the formation of an ionic chemical bond between lithium and fluorine atoms. Why is the resulting compou

valkas [14]

lithium and fluoride ions are arranged in an alternating three-dimensional crystal lattice. It's not possible to really isolate a single Li-F pair in the solid state. The many positive-negative electrical attractions hold the solid together. Ionic compounds are usually solids at room temperature since these electrical attractions are quite strong.

4 0

2 years ago

A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur

Ymorist [56]

Answer: The specific heat of metal is 0.821 J/g°C

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$