1. Sulfuric acid is formed when sulfur dioxide gas reacts with oxygen gas and water. Write a balanced chemical equation for the
1 answer:
Answer:
1. Balanced chemical for the given chemical reaction is :
number of mole of sulphuric acid produced = 12.5 mol
number of mole of oxygen required=6.25 mol
Explanation:
1. Balanced chemical for the given chemical reaction is :
from above balanced equation it is clearly that ,
2 mole of sulpher dioxide gives 2 mole of sulphuric acid
1 mole of sulpher dioxide gives 1 mole of sulphuric acid
therefore,
12 .5 mole of sulpher dioxide will givs 12.5 mole of sulphuric acid
number of mole of sulphuric acid produced = 12.5 mol
2 mole of sulphur dioxide needs 1 mole of oxygen gas
so,
1 mole of sulphur dioxide needs 0.5 mole of oxygen gas
therefore 12.5 of sulphur dioxide needs mole of oxygen
number of mole of oxygen required=6.25 mol
You might be interested in
Answer:
Option A and D are correct.
Unstable species react rapidly.
Stable species do not react rapidly.
Explanation:
The complete question is attached to this solution.
The more stable a reactant is, the less reactive it will be. A stable reactant has a very stable structure in which it will avoid any perturbations. And for a reaction to occur, the bonds in the reactant must break down to form the products. A stable reactant has very strong bonds that aren't easy to break down, hence, reactions involving very stable reactants do not proceed rapidly.
And the more unstable a reactant specie is, the more rapidly it reacts. This is why the reaction involving the less stable isotope of carbon; Carbon-14 is very rapid. It is the same reason as explained above that is responsible for this. The bond between unstable species are not strong and are easily breakable, thereby leading to a quick reaction.
Hope this Helps!!!
Answer:
1.44%
Explanation:
Initially, there is 24.201g of the waste. Let this waste contain 'x' mols of ammonia(). Now, this waste is dissolved in 72.053g of water, thus making the weight of the solution to 24.201+72.053 = 96.254g. Now, from this, 13.774g is taken. Note that, when the aqueous waste is dissolved in water, the ammonia particles are uniformly distributed, i.e, the 'x' mols of ammonia is uniformly present in the 96.254g of the solution. Hence, when we take 13.774g of the solution, only a fraction of the ammonia particles is taken. This fraction is equal to
of x, which is equal to 0.1431 times 'x'.
For the HCl solution, 28.02mL of 0.1045M solution contains 28.02x0.1045 = 2.9281mmols of HCl is present in it.
The basic titration reaction that occurs is : →
, i.e, one mol of ammonia requires one mol of HCl for neutralization. Therefore, for the above solution of HCl containing 2.9281mmols, same amount of ammonia, i.e, 2.9281mmols is required for complete neutralization.
Therefore, 0.1431x = 2.9281 mmols ⇒ x = 20.4619 mmols.
The molecular weight of ammonia is 17g/mol. Therefore, the weight of 20.4619mmols of ammonia has a weight(w) = 20.4619 x x 17 = 0.3478 g.
Therefore the weight of ammonia in the initial aqueous waste is 0.3478 g. The total weight of the waste is 24.201g, hence, the percent weight of ammonia is given by ×100 = 1.44%