How many moles of PCl5 can be produced from 53.0 g of Cl2 (and excess P4)?

Chemistry

1 answer:

elena55 [62]3 years ago

5 0

Answer:

0.299 moles of PCl5

Explanation:

First form an equation based of given information: P4 + Cl2= 4PCl5

Next balance it: P4 + 10Cl2 = 4PCl5

Take your given value (53.0g of Cl2) and divide it by its molar mass to get moles of Cl2: 53.0g/70.906g = 0.747 moles Cl2

Then, multiply by the molar ratio (10Cl2 to 4PCl5): (0.747 x 4 mol PCl5)/10 mol Cl2 = 0.299 moles PCl5

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If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$