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Snezhnost [94]
3 years ago
12

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

erature and pressure is
Chemistry
1 answer:
MArishka [77]3 years ago
3 0

Answer:

V = 11.2L are produced

Explanation:

... <em>Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..</em>

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

<em>Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.</em>

<em />

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

<em>Moles Al-Molar mass: 26.98g/mol-:</em>

27g * (1mol / 26.98g) = 1mol of Al

<em>Moles HCl:</em>

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and <em>HCl is limiting reactant.</em>

<em />

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

<em>Where V is volume</em>

<em>n are moles: 0.5mol</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K</em>

<em>P is pressure: 1atm.</em>

<em />

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>
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If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the theoretical yield of the reaction? What is the excess reac
Shkiper50 [21]

theoretical yield of ammonia (NH₃) = 121.38 g

The limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Explanation:

We have the following chemical reaction in which nitrogen react with hydrogen to produce ammonia:

N₂ + 3 H₂ → 2 NH₃

Now we need to calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

We see from the chemical reaction that 3 moles of H₂ react with 1 mole of N₂ so 50 moles of H₂ react with 16.67 moles of N₂ which is way more than the available N₂ quantity of 3.57 moles, so the limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).

Knowing this we devise the following reasoning:

if                1 mole of N₂ produces 2 moles of NH₃

then   3.57 moles of N₂ produces X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

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mass of NH₃ = 7.14 × 17 = 121.38 g (theoretical yield)

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7 0
3 years ago
Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

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For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

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