aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide
a. balance the chem eqn for this rxn
Al2S3+6H2O = 2Al(OH)3+3H2S
b.how many g of water are needed in the rxn to produce al(oh)3?
what is the formula i use here??
c. how many g of al(oh)3 are obtained from 10.5 mg of al2s?
mw of al(oh)3= 78 g/mol
10.5g to g= 0.0105
idk how to get the wt bcos idk the no. of moles :( hoow?
d. how many g of h2s are formed in the rxn?
mw of h2s=34 g/mol.. again idk how to get g :(
e. does the soln to the prblem obey 'law of conservation of mass'? why?
yes i think so cos the energy is being passed only but can never be created or destroyed. ?
b. To determine the amount of water needed to produce Al(OH)3, you need to use the balanced chemical equation. According to the equation:
Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
The stoichiometric ratio between water (H2O) and aluminum hydroxide (Al(OH)3) is 6:2. This means that for every 6 moles of water, 2 moles of Al(OH)3 are formed.
To find the amount of water in grams needed to produce a certain amount of Al(OH)3, you will need the molar mass of Al(OH)3, which is 78 g/mol.
c. To determine the amount of Al(OH)3 obtained from 10.5 mg of Al2S3, you need to calculate the number of moles of Al2S3 first.
Given:
Mass of Al2S3 = 10.5 mg = 0.0105 g
Molar mass of Al2S3 = 150 g/mol (2*27 + 3*32)
To find the number of moles, you can use the formula:
moles = mass / molar mass
moles of Al2S3 = 0.0105 g / 150 g/mol = 0.00007 mol
According to the balanced equation, 1 mole of Al2S3 produces 2 moles of Al(OH)3. Therefore, the number of moles of Al(OH)3 formed would be:
moles of Al(OH)3 = 2 * 0.00007 mol = 0.00014 mol
To convert moles to grams, you can use the formula:
grams = moles * molar mass
grams of Al(OH)3 = 0.00014 mol * 78 g/mol = 0.01092 g
Therefore, 0.01092 g of Al(OH)3 will be obtained from 10.5 mg of Al2S3.
d. To determine the amount of H2S formed in the reaction, you need to use the balanced chemical equation. According to the equation:
Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
The stoichiometric ratio between Al2S3 and H2S is 1:3. This means that for every 1 mole of Al2S3, 3 moles of H2S are formed.
Since you have already calculated the moles of Al2S3 in part c to be 0.00007 mol, the moles of H2S formed would be:
moles of H2S = 3 * 0.00007 mol = 0.00021 mol
To convert moles to grams, you can use the formula:
grams = moles * molar mass
molar mass of H2S = 34 g/mol
grams of H2S = 0.00021 mol * 34 g/mol = 0.00714 g
Therefore, 0.00714 g of H2S will be formed in the reaction.
e. Yes, the solution to the problem obeys the 'law of conservation of mass.' According to this law, mass cannot be created or destroyed in a chemical reaction. The total mass of the reactants is equal to the total mass of the products. In this case, the mass of the reactants (Al2S3 and H2O) is equal to the mass of the products (Al(OH)3 and H2S), so the law of conservation of mass is satisfied.
a. To balance the chemical equation, you need to ensure that the same number of atoms of each element is present on both sides of the equation. Here's the balanced equation:
Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
b. The formula you can use to determine the number of grams of water needed is:
Mass (g) = Number of moles × Molecular weight
In this case, you need to find the moles of Al(OH)3 formed and then multiply it by the molecular weight of water.
c. To determine the grams of Al(OH)3 obtained from 10.5 mg of Al2S, you can follow these steps:
1. Convert 10.5 mg to grams: 10.5 mg = 0.0105 g
2. Calculate the number of moles of Al2S3: moles = mass / molecular weight
3. Use the balanced equation to relate the moles of Al2S3 to moles of Al(OH)3.
4. Convert moles of Al(OH)3 to grams using the formula mentioned in part b.
d. To find the grams of H2S formed, you can follow a similar process as in part c, using the molecular weight of H2S.
e. Yes, the solution obeys the law of conservation of mass. According to this law, mass is neither created nor destroyed in a chemical reaction, it is only rearranged. In this reaction, the total mass of the reactants (Aluminum sulfide and water) is equal to the total mass of the products (Aluminum hydroxide and Hydrogen sulfide).
aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide
a. balance the chem eqn for this rxn
Al2S3+6H2O = 2Al(OH)3+3H2S
right.
b.how many g of water are needed in the rxn to produce al(oh)3?
what is the formula i use here??
You can't do without knowint how much Al(OH)3 you want to produce? By the way, you need to find the caps key and use it.
c. how many g of al(oh)3 are obtained from 10.5 mg of al2s?
mw of al(oh)3= 78 g/mol
10.5g to g= 0.0105
idk how to get the wt bcos idk the no. of moles :( hoow?
1.Find mols Al2S3. mols Al2S3 = grams/molar mass = 0.0105/molar mass Al2S3.
2. Convert mols of anything to mols anything by using the coefficients in the balanced equation. mols Al2S3 x (2 mols Al(OH)3/1 mol Al2S3) = mols Al2S3 x 2/1 = ?
3. Convert mols Al(OH)3 to grams by grams Al(OH)3 = mols Al(OH)3 x molar mass Al(OH)3.
d. how many g of h2s are formed in the rxn?
mw of h2s=34 g/mol.. again idk how to get g :(
Done the same way as converting grams Al2S3 to grams Al(OH)3.
e. does the soln to the prblem obey 'law of conservation of mass'? why?
yes i think so cos the energy is being passed only but can never be created or destroyed. ?
Your answer of yes is correct but your reasoning is lousy (you just restated the law). To show it does obey the Law of Conservation of mass add mass of the reactants together and mass of products together. mass reactants and mass of products shold be the same.