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Mademuasel [1]
3 years ago
12

How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in th

e reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
Chemistry
2 answers:
vitfil [10]3 years ago
7 0
How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
Sergeeva-Olga [200]3 years ago
5 0

<u>Answer:</u> The mass of hydrogen gas produced by the reaction is 6.9 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of aluminium = 63 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{63g}{27g/mol}=2.33mol

For the given chemical reaction:

2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)

Hydrochloric acid is present in excess. So, it is considered as an excess reagent. And, aluminium metal is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 moles of hydrogen gas.

So, 2.33 moles of aluminium metal will produce = \frac{3}{2}\times 2.33=3.45mol of hydrogen gas

Now, calculating the mass of hydrogen gas by using equation 1:

Moles of hydrogen gas = 3.45 moles

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

3.45mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(3.45mol\times 2g/mol)=6.9g

Hence, the mass of hydrogen gas produced by the reaction is 6.9 grams

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Determine the molar concentration of na+ and po4 3- in a 2.25 M Na3 PO4 solution
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Answer:

A. The concentration of Na^+ in the solution is 6.75 M.

B. The concentration of PO4^3- in the solution is 2.25 M.

Explanation:

We'll begin by writing the balanced dissociation equation for Na3PO4.

This is illustrated below:

Na3PO4 will dissociate in solution as follow:

Na3PO4(aq) —> 3Na^+(aq) + PO4^3-(aq)

Thus, from the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+ and 1 mole of PO4^3-

A. Determination of the concentration of Na+ in 2.25 M Na3PO4 solution.

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From the balanced equation above,

1 mole of Na3PO4 produce 3 moles of Na^+.

Therefore, 2.25 M Na3PO4 solution will produce = (2.25 x 3) /1 = 6.75 M Na^+.

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This can be obtained as follow:

From the balanced equation above,

1 mole of Na3PO4 produce 1 mole of PO4^3-

Therefore, 2.25 M Na3PO4 solution will also produce 2.25 M PO4^3-.

Therefore, the concentration of PO4^3- in the solution is 2.25 M.

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