All categories

3 years ago

Limiting Reactant

Chemistry

1 answer:

liraira [26]3 years ago

3 0

Answer:

(1) Cl₂ is the limiting reactant.

(2) 8.18 g

Explanation:

2Na(s) + Cl₂(g) → 2NaCl(s)

First we convert the given masses of reactants into moles, using their respective molar masses:

Na ⇒ 12.0 g ÷ 23 g/mol = 0.522 mol Na

Cl₂ ⇒ 5.00 g ÷ 70.9 g/mol = 0.070 mol Cl₂

0.070 moles of Cl₂ would react completely with (2 * 0.070) 0.14 moles of Na. There are more Na moles than that, so Na is the reactant in excess while Cl₂ is the limiting reactant.

Then we calculate how many moles of NaCl are formed, using the limiting reactant:

0.070 mol Cl₂ * $\frac{2molNaCl}{1molCl_2}$ = 0.14 mol NaCl

Finally we convert NaCl moles into grams:

0.14 mol NaCl * 58.44 g/mol = 8.18 g

You might be interested in

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

3 years ago

According to Le Chatelier’s principle, what always happens to the equilibrium of a reaction when the temperature is reduced?

Tasya [4]

It always shift to the direction where balance out the reaction
here
It shifts in the exothermic direction.

5 0

3 years ago

Read 2 more answers

Which of the following statements is true about catalysts? and Why?

LUCKY_DIMON [66]

I believe the correct answer from the choices listed above is option D. Catalysts lower the activation energy of a chemical reaction. It is a substance which speeds up a reaction, but is chemically unchanged at the end of the reaction. It provides another pathway for the reaction to occur.

8 0

4 years ago

Read 2 more answers

According to rounding rules for addition the sum of 27.1, 34.538, and 37.68 is

Katena32 [7]

The answer should be...99.318!

5 0

3 years ago

Why are ions usually formed during the interaction of metal with nonmetal and not between two metals?

Alik [6]

Explanation:

Ions are always formed when metals and non-metals interact because metals are electropositive. They willing release electrons to non-metals that are electronegative.

This activity results in charge separation. The transfer of electrons from one specie to another is what results in an ionic bond and the precedence of charged particles.

Between non-metals, the electrons are jointly shared. Therefore, there is no charge separation.

7 0

3 years ago