In this case, considering the widely studied Le Chatelier's principle, we can realize that the factors affecting equilibrium are concentration, temperature and pressure and volume if the reaction is in gaseous phase and with non-zero change in the number of moles. In such a way, by adding a catalyst to given reaction will have no effect on the equilibrium direction.

Best regards.

7 0

3 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

6 0

3 years ago

If a black hole is invisible how can we determine when one is present

Gelneren [198K]

Answer: you can’t

Explanation: i’m dumb too

8 0

3 years ago