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3 years ago

How many moles are present in 102.55g of water (H20)

Chemistry

2 answers:

natulia [17]3 years ago

6 0

About 5 moles of water are present. To be precise it’s 5.666666667 moles of H2O

ElenaW [278]3 years ago

3 0

Answer: 5.70 moles H2O

Explanation: 1 mole of H2O is equal to its molar mass which is 18 g.

102.55 g H2O x 1mole H2O / 18 g H2O

= 5.70 moles H2O

Note: the answer is rounded off.

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2 years ago

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If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

Luba_88 [7]

<u>Answer:</u> The value of $K_c$ is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

The expression for the $K_{c}$ is given as:

$K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

We are given:

$[NH_3]=0.25M$

$[H_2]=0.75M$

$[N_2]=1.1M$

Putting values in above equation, we get:

$K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}$

$K_c=0.135$

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the given reaction, the value of $K_c$ is less than 1. Thus, the reaction is reactant favored.

Hence, the value of $K_c$ is 0.136 and is reactant favored.

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3 years ago