I’ve tried everything but just can’t get it.
Write the reaction between solutions of mercury (II) nitrate and sodium sulfide.
a) how many moles of sodium nitrate form from the reaction of 2.85 mol of sodium sulfide?
b) how many litres of 0.150 mol/L mercury (II) nitrate react with 0.540 L of 0.653 mol/L sodium sulfide?
c) how many grams of precipitate would you get from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate?
Write the reaction between solutions of mercury (II) nitrate and sodium sulfide.
Hg(NO3)2 + Na2S ==> HgS + 2NaNO3
a) how many moles of sodium nitrate form from the reaction of 2.85 mol of sodium sulfide?
2.85 mols Na2S x (2 mols NaNO3/1 mol Na2S) = ? mols NaNO3
b) how many litres of 0.150 mol/L mercury (II) nitrate react with 0.540 L of 0.653 mol/L sodium sulfide?
mols Na2S = M x L = 0.653 M x 0.540 L = ?
mols Hg(NO3)2 = mols Na2S. Same reasoning as part a.
Then M Hg(NO3)2 = mols/L. You know mols and M, solve for L
c) how many grams of precipitate would you get from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate?
mols Na2S = M x L = 0.550 x 1.00 L = 0.550 mols.
You get 1 mol HgS(s) for every mol Na2S so mols HgS = mols Na2S using the same reasoning as part a.
Then grams = mols HgS x molar mass = ? You know mols HgS and molar mass HgS so substitute and solve for grams.
Post work if you get stuck. Be specific if you don't understand one of the steps above.
The reaction between solutions of mercury (II) nitrate and sodium sulfide can be represented as follows:
Hg(NO3)2(aq) + Na2S(aq) -> HgS(s) + 2NaNO3(aq)
a) To determine the number of moles of sodium nitrate formed from the reaction of 2.85 mol of sodium sulfide, we refer to the balanced chemical equation. From the equation, we can see that the stoichiometric ratio between sodium sulfide and sodium nitrate is 1:2. Therefore, for every 1 mole of sodium sulfide, 2 moles of sodium nitrate are formed. Hence, the number of moles of sodium nitrate formed from 2.85 mol of sodium sulfide would be:
2.85 mol sodium sulfide * (2 mol sodium nitrate / 1 mol sodium sulfide) = 5.7 mol sodium nitrate
b) To determine the volume of 0.150 mol/L mercury (II) nitrate required to react with 0.540 L of 0.653 mol/L sodium sulfide, we can use the stoichiometric ratio from the balanced chemical equation. The ratio between mercury (II) nitrate and sodium sulfide is 1:1. Therefore, the mole ratio is 1:1.
From the given information, we have:
Volume of sodium sulfide = 0.540 L
Concentration of sodium sulfide = 0.653 mol/L
Mole ratio between mercury (II) nitrate and sodium sulfide = 1:1
Using the formula: Moles = Concentration * Volume, we can calculate the number of moles of sodium sulfide:
Moles of sodium sulfide = 0.653 mol/L * 0.540 L = 0.35202 mol
Since the mole ratio is 1:1, we would require the same number of moles of mercury (II) nitrate. Therefore, the volume of 0.150 mol/L mercury (II) nitrate would be:
Volume of mercury (II) nitrate = Moles of mercury (II) nitrate / Concentration of mercury (II) nitrate
Volume of mercury (II) nitrate = 0.35202 mol / 0.150 mol/L = 2.3468 L (rounded to 4 decimal places)
So, 2.3468 L of 0.150 mol/L mercury (II) nitrate would react with 0.540 L of 0.653 mol/L sodium sulfide.
c) To determine the mass of precipitate formed from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate, we need to use stoichiometry.
From the balanced chemical equation, we can see that the stoichiometric ratio between sodium sulfide and mercury(II) sulfide is 1:1. Therefore, for every 1 mole of sodium sulfide, 1 mole of mercury(II) sulfide (precipitate) is formed.
The number of moles of sodium sulfide is given by:
Moles of sodium sulfide = Concentration * Volume
Moles of sodium sulfide = 0.550 mol/L * 1.00 L = 0.550 mol
Since the stoichiometric ratio is 1:1, the number of moles of mercury(II) sulfide formed would also be 0.550 mol.
To calculate the mass of mercury(II) sulfide precipitate, we need to know its molar mass. The molar mass of mercury(II) sulfide (HgS) is 200.59 g/mol.
Mass of mercury(II) sulfide precipitate = Moles of mercury(II) sulfide * Molar mass
Mass of mercury(II) sulfide precipitate = 0.550 mol * 200.59 g/mol = 110.32 g (rounded to 2 decimal places)
Therefore, from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate, you would obtain 110.32 grams of precipitate (mercury(II) sulfide).
To determine the reaction between mercury (II) nitrate and sodium sulfide, we need to first write the balanced chemical equation.
The chemical equation for the reaction is:
Hg(NO3)2 + Na2S → HgS + 2 NaNO3
Now that we have the balanced equation, we can proceed to answer the questions:
a) To find the number of moles of sodium nitrate formed from the reaction of 2.85 mol of sodium sulfide, we need to use the stoichiometry of the balanced equation. From the equation, we can see that for every 1 mole of sodium sulfide, 2 moles of sodium nitrate are formed.
Therefore, if we start with 2.85 mol of sodium sulfide, we would get twice as many moles of sodium nitrate. So, the number of moles of sodium nitrate formed would be 2 * 2.85 mol = 5.70 mol.
b) To determine the volume of 0.150 mol/L mercury (II) nitrate that reacts with 0.540 L of 0.653 mol/L sodium sulfide, we need to use the stoichiometry of the balanced equation.
From the equation, we can see that the molar ratio between mercury (II) nitrate and sodium sulfide is 1:1. This means that one mole of mercury (II) nitrate reacts with one mole of sodium sulfide.
First, we need to find the number of moles of mercury (II) nitrate present in the given volume using concentration and volume relationship:
moles of mercury (II) nitrate = concentration * volume
moles of mercury (II) nitrate = 0.150 mol/L * 0.540 L = 0.081 mol
Since the molar ratio is 1:1, the 0.081 mol of mercury (II) nitrate reacts with 0.081 mol of sodium sulfide.
c) To find the mass of precipitate formed from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate, we again need to use the balanced equation and stoichiometry.
From the balanced equation, we can see that the molar ratio between sodium sulfide and mercury (II) nitrate is 1:1. This means that one mole of sodium sulfide reacts with one mole of mercury (II) nitrate.
First, we need to find the number of moles of sodium sulfide present in the given volume using concentration and volume relationship:
moles of sodium sulfide = concentration * volume
moles of sodium sulfide = 0.550 mol/L * 1.00 L = 0.550 mol
Since the molar ratio is 1:1, the 0.550 mol of sodium sulfide reacts with 0.550 mol of mercury (II) nitrate.
To calculate the mass of the precipitate (HgS) formed, we need to know the molar mass of HgS:
molar mass of HgS = molar mass of Hg + molar mass of S = 200.61 g/mol + 32.07 g/mol = 232.68 g/mol
The mass of the precipitate can be calculated by multiplying the number of moles of HgS by its molar mass:
mass = moles * molar mass
mass = 0.550 mol * 232.68 g/mol = 127.97 g
Therefore, the mass of the precipitate formed would be approximately 127.97 grams.