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bija089 [108]
3 years ago
10

When 83.3 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, h

ow many moles of hydrochloric acid will be produced?
PCl5 + H2O yields H3PO4 + HCl

0.4 moles
1.0 moles
1.6 moles
2.0 moles
Chemistry
2 answers:
Lelechka [254]3 years ago
8 0

Answer:

The answer is 2 moles of HCl

Explanation:

To solve we have to adjust the equation:

PCl5 + 4H2O = 5HCl + H3PO4

we calculate the molecular weights of the compounds using the periodic table:

P = 30.97 g/mol

Cl = 5x35.45 = 177.25 g/mol

H = 1 g/mol

PCl5 = 30.97+177.25 = 208.22 g/mol

HCl =1+35.45 = 36.45 g/mol

we use a rule of three to calculate the mass of hydrochloric acid that forms when the 83.3 phosphorus pentachloride reacts:

208.22 g PCl5-----------------182.25 g HCl

83.3 g PCl5------------------------X g HCl

Clearing the X, we have:

X g HCl = (83.3x182.25)/208.22 = 72.91 g HCl

we calculate the moles of hydrochloric acid:

Moles of HCl = 72.91/36.45 = 2 moles of HCl

Nataly [62]3 years ago
6 0
<span>2.0 moles. (83.3g / 208 [grams/moles] *5 [from balancing] = around 2 </span>
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PtichkaEL [24]

Answer:

Option D. 30 g

Explanation:

The balanced equation for the reaction is given below:

2Na + S —> Na₂S

Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of S = 32 g/mol

Mass of S from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted with 32 g of S.

Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.

Thus, 30 g of S is needed for the reaction.

8 0
2 years ago
a chemistry teacher adds 50.0 ml of 1.50 m h2so4 solution to 200 ml of water. What is the concentration of the final solution?
nevsk [136]

Answer:0.300M

Explanation:1) Data:

a) Initial solution

M = 1.50M

V = 50.0 ml = 0.050 l

b) Solvent added = 200 ml = 0.200 l

2) Formula:

Molarity: M = moles of solute / volume of solution is liters

3) Solution:

a) initial solution:

Clearing moles from the molarity formula: moles = M × V

moles of H₂SO₄ = M × V = 1.5M × 0.050 l = 0.075 mol

b) final solution:

i) Volumen of solution = 0.050 l + 0.200l = 0.250l

ii) M = 0.075 mol / 0.250 l = 0.300M ← answeer

5 0
2 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

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vitfil [10]

Answer:

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Explanation:

Calculate the mass

Let x be the mass

If 1 mol of Fe has a mass of 55.85 g

1 mol =55.85g

122mol= x

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