All categories

3 years ago

Predict the effect of an eightfold pressure increase on the equilibriunm composition of the reaction 3 Nalg) + H21g) =2 NaHlg)

2 answers:

6 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

Conclusion- the reaction will shift to the right of the reaction

Assoli18 [71]3 years ago

4 0

According to Le cha telier's principle when we increase the pressure of a equilibrium reaction the reaction shift to the side where few moles of gas present.

Your reaction (Notice - I guess the reaction you written is wrong, but still I'm solving with your given prediction)

3NaI (g) + H2 (g) = 2NaHI (g)

Where access of sodium iodide is reacting with Hydrogen gas to form NaHI molecule.

Number of moles of gas on reaction side - 3+2 = 5 moles

Number of moles of gas on product side - 2 moles

Conclusion- the reaction will shift to the right of the reaction

You might be interested in

Calculate the number of grams of CH_3COOH in the vinegar.

AysviL [449]

when density in g/ml
9.86ml CH3COOH (X g/1ml) = g
1.049g/ml
9.86ml (1.049g/1ml) = 10.343 g
hope it helps

6 0

4 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1