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3 years ago

Write the concentration equilibrium constant expression for this reaction.

1 answer:

8 0

In a chemical reaction, the equilibrium constant refers to the value of its reaction quotient at chemical equilibrium, that is, a condition attained by a dynamic chemical system after adequate time has passed, and at which its composition has no measurable capacity to undergo any kind of further modification.

The given reaction is: HCN (aq) + OH⁻ = CN⁻ (aq) + H2O (l)

The equilibrium constant = product of concentration of products / product of concentration of reactants

(Here, H2O is not considered as its concentration is very high)

So, Keq = [CN⁻] / [HCN] [OH⁻]

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If 3.0 g of NH3 reacts with 5.0 g of HCl, what is the limiting reactant?

ELEN [110]

<h2>Answer:HCl</h2>

Explanation:

$\text{number of moles}=\frac{\text{given mass}}{\text{molar mass}}$

For $NH_{3},$

$\text{given mass}=3g$

$\text{molar mass}=14+3=17g$

$n_{NH_{3}}=\frac{3}{17}=0.176$

For $HCl,$

$\text{given mass}=5g$

$\text{molar mass}=1+35.5=36.5g$

$n_{HCl}=\frac{5}{36.5}=0.136$

The reaction between $NH_{3}$ and $HCl$ is

$NH_{3}+HCl$ → $NH_{4}Cl$

So,one mole of $NH_{3}$ requires one mole of $HCl$

$0.176$ moles of $NH_{3}$ requires $0.176$ moles of $HCl$

But there are only $0.136$ moles of $HCl$ available.

$HCl$ will be consumed first.

So, $HCl$ is the limiting reagent.

7 0

3 years ago

Theoretically how many moles of carbonic acid will be produced by the 3.00 g sample of NaHCO3

yaroslaw [1]

NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3

</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>

4 0

3 years ago

Read 2 more answers

A piece of metal with a mass of 147g and a volume of 21ml. Calculate the density of the metal.

kipiarov [429]

Answer:

density of a piece of metal = 7 gr/ml

Explanation:

See the file please