Sodium(Na) is the limiting reagent.
<h3>What is Limiting reagent?</h3>
The reactant that is totally consumed during a reaction, or the limiting reagent, decides when the process comes to an end. The precise quantity of reactant required to react with another element may be estimated from the reaction stoichiometry.
How do you identify a limiting reagent?
The limiting reactant is the one that is consumed first and sets a limit on the quantity of product(s) that can be produced. Calculate how many moles of each reactant are present and contrast this ratio with the mole ratio of the reactants in the balanced chemical equation to get the limiting reactant.
Start by writing the balanced chemical equation that describes this reaction
Notice that the reaction consumes 2 moles of sodium metal for every 1 mole of chlorine gas that takes part in the reaction and produces 2 moles of sodium chloride.
now we can see that we have 3 moles of sodium and 3 moles of chlorine, according to question. so, we can say that sodium is the limiting reagent in the given situation.
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The change in internal energy of the combustion of biphenyl in Kj is calculated as follows
=heat capacity of bomb calorimeter x delta T where delta T is change in temperature
delta T = 29.4 -25.8= 3.6 c
= 5.86 kj/c x 3.6 c = 21.096 kj
Answer:
Explanation:
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In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:
It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:
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Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
