How many milliliters of 2.00 M H2SO4 will react with 28.0 g of NaOH?

I just want to check to see if the answer my child has is right.

Law Incorporation [45]

Answer:

2= Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol

3= Mass of hydrogen = 0.45 g

3= Mass of oxygen = 3.616 g

4 = Theoretical yield of P₂O₅ = 14.2 g.

Explanation:

Q2= Given data:

Moles of PbS = 0.356 mol

Moles of PbO = ?

Moles of SO₂ = ?

Solution:

Chemical equation:

2PbS + 3O₂ → 2PbO + 2SO₂

Now we will compare the moles of PbS with PbO and SO₂ from balanced chemical equation.

PbS : PbO

2 : 2

0.356 : 0.356

Moles of PbO produced from 0.356 moles of PbS = 0.356 mol

PbS : SO₂

2 : 2

0.356 : 0.356

Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol

Q= 3

Given data:

Mass of water = 4.05 g

Mass of hydrogen = ?

Mass of oxygen = ?

Solution:

Chemical equation:

2H₂O → 2H₂ + O₂

Number of moles of water:

Number of moles of water = mass/ molar mass

Number of moles of water = 4.05 g/ 18 g/mol

Number of moles of water = 0.225 mol

Now we compare the moles of water with hydrogen and oxygen.

H₂O : H₂

2 : 2

0.225 : 0.225

Mass of hydrogen:

Mass of hydrogen = moles × molar mass

Mass of hydrogen = 0.225 mol × 2g/mol

Mass of hydrogen = 0.45 g

H₂O : O₂

2 : 1

0.225 : 1/2×0.225 = 0.113 mol

Mass of oxygen:

Mass of oxygen = moles × molar mass

Mass of oxygen = 0.113 mol × 32g/mol

Mass of oxygen = 3.616 g

Q 4

Given data:

Mass of phosphorus = 3.07 g

Mass of oxygen = 6.09 g

Theoretical yield of P₂O₅ = ?

Chemical equation:

4P + 5O₂ → 2P₂O₅

Number of moles of phosphorus = mass/ molar mass

Number of moles of phosphorus = 3.07 g/ 31 g/mol

Number of moles of phosphorus = 0.1 mol

Number of moles of oxygen = mass/ molar mass

Number of moles of oxygen = 6.09 g/ 32 g/mol

Number of moles of oxygen = 0.2 mol

Now we will compare the moles of P₂O₅ with oxygen and phosphorus.

O₂ : P₂O₅

5 : 2

0.2 : 2/5 ×0.2 = 0.08 mol

P : P₂O₅

4 : 2

0.1 : 2/4×0.1 = 0.05 mol

The number of moles of P₂O₅ produced from phosphorus are less that's why phosphorus will be limiting reactant.

Mass of P₂O₅ = moles × molar mass

Mass of P₂O₅ = 0.05 mol × 283.88 g/mol

Mass of P₂O₅ = 14.2 g

Theoretical yield of P₂O₅ = 14.2 g.

3 0

4 years ago

Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl ,

puteri [66]

Answer:

See explanation

Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

8 0

3 years ago

The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 218 kJ/mol and the change in enthalpy for the

madreJ [45]

Answer:

$470\frac{KJ}{mol}$

Explanation:

The activation energy represents the energy barrier that reagents must pass to transform into products (or products to transform into reagents in a reverse reaction)

For any reaction, the change in enthalpy is related to the activation energy by the equation

$\Delta H =E_{a\ direct}-E_{a\ reverse}$

So, the activation energy for the reverse reaction is

$E_{a\ reverse}=E_{a\ direct}-\Delta H =218 \frac{KJ}{mol} - (-252)\frac{KJ}{mol}=470\frac{KJ}{mol}$

4 0

3 years ago

Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10−11. It is used to contr

VLD [36.1K]

Answer:

1.27 x 10⁻⁴M

Explanation:

For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.

Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³

Solve for 'x' => x = Solubility = ∛Ksp/4

4 0

4 years ago

What ions and/or molecules (apart from water) are present in relatively large proportions in a solution of a weak acid hclo (aq)

matrenka [14]

HClO, perchloric acid is a weak acid. Unlike strong acids like HCl or H2SO4 it not dissociate completely but partially such that an equilibrium exists between the dissociated ions and the undissociated acid. The equilibrium is as shown below:

HClO + H2O ↔ H3O⁺ + ClO⁻

Since HClO is a weak acid, the reverse reaction is favored over the forward reaction. Thus apart from water, HClO will be present in large amounts.

3 0

3 years ago