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3 years ago

6.0 moles of Na and 4.0 moles of Cl2 are mixed, how manu moles of NaCl in moles cane be made from this mixture

Chemistry

1 answer:

Sladkaya [172]3 years ago

3 0

Answer:

Moles of NaCl formed is 6.0 moles

Explanation:

We are given the equation;

2 Na(s) + Cl₂(g) → 2 NaCl(s)

Moles of Na is 6.0 moles
Moles of Cl₂ is 4.0 moles

From the reaction;

2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl

In this case;

6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.

Thus, the rate limiting reagent is sodium.

But, 2 moles of sodium reacts to form 2 moles of NaCl

Therefore;

Moles of NaCl = Moles of Na

= 6.0 moles

Thus, moles of sodium chloride produced is 6.0 moles

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Answer:

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Explanation:

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2 years ago

A solution is made by adding 20 g table salt to 100 mL water. The solubility of salt is 36 g/100 mL water. Which term best descr

marysya [2.9K]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

N204(0) + 2NO2(g)

user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

$\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}$

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state

Q is the product of the ion ions from the reacting substance

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

$\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}$