All categories

3 years ago

22.17 grams of Mercury (II) Nitrate, Hg(NO3)2, reacts with an excess of Potassium, K.

Chemistry

1 answer:

Svetradugi [14.3K]3 years ago

7 0

Answer:

Mass = 14.0 g

Explanation:

Given data:

Mass of mercury nitrate = 22.17 g

Mass of mercury formed = ?

Solution:

Chemical equation:

Hg(NO₃)₂ + 2K → 2KNO₃ + Hg

Number of moles of mercury nitrate:

Number of moles = mass/molar mass

Number of moles = 22.17 g / 324.6 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of Hg(NO₃)₂ and mercury.

Hg(NO₃)₂ : Hg

1 : 1

0.07 : 0.07

Mass of mercury:

Mass = number of moles × molar mass

Mass = 0.07 mol × 200.6 g/mol

Mass = 14.0 g

You might be interested in

Identify the parts of the telescope.

Alona [7]

Eyepiece, finder-scope, optical tube, aperture, focuser, and mount

6 0

3 years ago

Read 2 more answers

What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)

Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

$Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\ ml\\\\V = 33.33 \ ml$

Therefore, volume required is 33.33 ml .

5 0

3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$