Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL

We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;

Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;

The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;

The number of grams of aluminum required to react with HCl is 0.6258 g.
Answer:
Explanation:
Your answer should be in the attached pdf
When a bath bomb comes in contact with water, the baking soda and citric acid react to make carbon dioxide bubbles. This is an acid–base reaction, where baking soda (also called sodium bicarbonate) is a weak base and citric acid is a weak acid.
Answer:
solubility of X in water at 17.0
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0
can be calculated using the information given.
Let's assume solubility of X in water at 17.0
is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y =
= 0.11
Hence solubility of X in water at 17.0
is 0.11 g/mL.