Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to

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For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

Answer:

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

Where

is the number of moles of hydrogen

is the mass of hydrogen

is the density of hydrogen

One electron is involved. Explanation: In redox reactions, determining the equivalents requires knowledge of the number of transferred electrons. In this specific case, one equivalent corresponds to a transfer of a single electron.

The slight warm feeling noticed at the valve stem when air is pumped into the tire is likely due to the kinetic energy generated by the friction from the pump and the resultant increase in gas pressure within the tire.

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %