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11111nata11111 [884]

3 years ago

12

In the reaction 2H2+____O2----2H2O, what coefficient should be placed in front of O2 to balance the reaction? Please help

1 answer:

Natali5045456 [20]3 years ago

7 0

Answer:

D

Explanation:

2H2+______O2- - - - - 2H2O

I'd say no coefficient is needed cos it's already balanced

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2 years ago

Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce

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The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

$\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }$

In the given acid nitric:

Owing to the nitric acid mass = $75.9 g$

Nitric acid molar weight $= 63\ \frac{g}{mol}$

If they put values above the formula, they receive:

$\text{moles in nitric acid} = \frac{75.9}{63}$

$=1.204 \ mol$

In the given chemical equation:

$3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)$

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

$= \frac{1}{2} \times 1.204$

$=0.602\ \ \text{mol of water}.$

We are now using Equation 1 in determining the quantity of water:

Water moles $= 0.602\ mol$

Water weight molar $= 18.02 \ \frac{g}{mol}$

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3 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

<u />

<u>4. Equilibrium expression</u>

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

<u />

<u>5. Solve:</u>

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago