Answer:
144g of H₂O
Explanation:
3NH₄ClO₄(s) + 3Al → Al₂O₃(s) + AlCl₃(s) + 3NO(g) + 6H₂O(g)
From the equation:
3 moles of NH₄ClO₄ produced 6 moles of H₂O
4 moles of NH₄ClO₄ produced ? moles of H₂O
(4 ₓ 6)/3 = = 8 moles of H₂O
1 mole of H₂O = (1 × 2) + 16 = 18g (The Relative Molecular mass of H₂O)
8 moles of H₂O = ?
Therefore 8 × 18 = 144g
=144g of H₂O
Answer : The reagent present in excess and remains unreacted is,
Solution : Given,
Moles of = 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Hence, the reagent present in excess and remains unreacted is,
Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide,
Explanation:
The reaction equation for given reaction is as follows.
Here, 1 mole of reacts with 3 moles of
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of is calculated as follows.
Now, moles of .given by 0.5 mol of
is calculated as follows.
As molar mass of is 2.016 g/mol. Therefore, mass of
is calculated as follows.
Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, .
Answer: 1.32
Explanation:
First, we must obtain the molar mass of HBr. After that, we try to obtain the concentration of the hydrobromic acid from the formula n=CV since the volume of solution and mass of acid was provided. Recall that n=m/M. If the concentration of acid is thus obtained, we make use of the fact that the concentration of H+ in the acid is equal to the molar concentration of HBr to obtain the pH. The pH is the negative logarithm of the concentration we obtained in the initial step.
