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nadezda [96]
3 years ago
14

How many grams of product might form for the following reaction if 33.8 L of Oxygen gas is used in the following reaction? LiCl

+ O2 -> LiClO3
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

91.41 g of LiClO₃.

Explanation:

We'll begin by calculating the number of mole of O₂ that occupied 33.8 L. This can be obtained as follow:

22.4 L = 1 mole of O₂

Therefore,

33.8 L = 33.8 L × 1 mole / 22.4 L

33.8 L = 1.51 mole of O₂

Next, the balanced equation for the reaction.

2LiCl + 3O₂ —> 2LiClO₃

From the balanced equation above,

3 moles of O₂ reacted to produce 2 moles of LiClO₃.

Therefore, 1.51 mole of O₂ will react to produce = (1.51 × 2)/3 = 1.01 mole of LiClO₃.

Finally, we shall determine the mass of 1.01 mole of LiClO₃. This can be obtained as follow:

Mole of LiClO₃ = 1.01 mole

Molar mass of LiClO₃ = 7 + 35.5 + (3×16)

= 7 + 35.5 + 48

= 90.5 g/mol

Mass of LiClO₃ =?

Mass = mole × molar mass

Mass of LiClO₃ = 1.01 × 90.5

Mass of LiClO₃ = 91.41 g

Thus, 91.41 g of LiClO₃ were obtained from the reaction.

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5 0
3 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
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Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

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k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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