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4 years ago

Consider the chemical equations shown here.

Chemistry

1 answer:

jarptica [38.1K]4 years ago

6 0

Answer:

-1,300 kJ

Explanation:

P₄ + 3 O₂ ⇒ P₄O₆ ΔH = -1,640.1 kJ (EQ 1)

P₄O₁₀ ⇒ P₄ + 5 O₂ ΔH = 2,940.1 kJ (EQ 2)

These are the equations you are given. You need to make these equations into the equation below.

P₄O₆ + 2 O₂ ⇒ P₄O₁₀ (end equation)

Look at the product side of the end equation. You need to produce P₄O₁₀. In the EQ 2, P₄O₁₀ is on the reactant side. Flip the equation. Since you flipped the equation, the enthalpy will have the opposite sign.

P₄ + 5 O₂ ⇒ P₄O₁₀ ΔH = -2,940.1 kJ

On the reactant side of the end equation, you need P₄O₆ and 2 O₂. First, rearrange the equation so that P₄O₆ is on the right side. In EQ 1, P₄O₆ is on the product side. Flip the equation. Like the last one, the sign will change.

Now, cancel out all possible values. P₄ will cancel out since there is one on each side of the equation. Since there is 5 O₂ on one side and 3 O₂ on the other, subtract the two and put the remainder on the side of the larger value.

P₄ + 5 O₂ ⇒ P₄O₁₀ ΔH = -2,940.1 kJ

P₄O₆ ⇒ P₄ + 3 O₂ ΔH = 1,640.1 kJ

P₄O₆ + 2 O₂ ⇒ P₄O₁₀

This should be the resulting equation. Now, add the two enthalpies together to find the overall enthalpy.

-2,940.1 kJ + 1,640.1 kJ = -1,300 kJ

The overall enthalpy is -1,300 kJ.

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the molecular weight of the nitrous oxide is 44.013 g/mol. assuming standard temperature and pressure, what would be the volume,

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The volume of the nitrous oxide gas is 1729.3 Liters

<h3>What is the number of moles of gas present in 3.40 kg of nitrous oxide?</h3>

The number of moles of gas present in 3.40 kg of nitrous oxide is determined from the formula below:

Numbers of moles = mass/molar mass

the mass of nitrous oxide = 3.40 kg or 3400 g

the molar mass of nitrous oxide = 44.013 g/mol

Moles of gas = 3400 / 44.013

Moles of gas = 77.25 moles

Using the ideal gas equation to determine the volume of the gas:

PV= nRT

V = nRT/P

where;

V is the volume of gas
n is the number of moles of gas
R is molar Gas constant = 0.082 L.atm/mol/K
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3 years ago

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19 grams of sodium and 34 grams of chlorine are measured out and put into separate glass vials that weigh 10 grams each. Then so

vaieri [72.5K]

Answer:

63.02 g.

Explanation:

Na reacts with Cl₂ according to the balanced equation:

2Na + Cl₂ → 2NaCl,

It is clear that 2 mole of Na react with 1 mole of Cl₂ to produce 2 moles of NaCl.

Firstly, we need to calculate the no. of moles of Na and Cl₂:

no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.

no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.

From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).

So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".

The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.

∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.

To get the no. of grams of produced NaCl:

using cross multiplication:

2 mol of Na produce → 2 mol of NaCl, from the stichiometry.

∴ 0.826 mol of Na produce → 0.826 mol of NaCl.

∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.

∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.

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