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pentagon [3]
4 years ago
8

Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s)+3H2SO4(

aq)→Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 14.6 gg . Part A What minimum mass of H2SO4H2SO4 would you need? Express your answer in grams.
Chemistry
2 answers:
yawa3891 [41]4 years ago
6 0

Answer:

We need 79.6grams of H2SO4

Explanation:

Step 1: data given

Mass of aluminium = 14.6 grams

Molar mass aluminium = 26.98 g/mol

Molar mass H2SO4 = 98.08 g/mol

Step 2: The balanced equation

2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

Step 3: Calculate moles Aluminium

Moles Aluminium = mass aluminium / molar mass aluminium

Moles aluminium = 14.6 grams / 26.98 g/mol

Moles aluminium = 0.541 moles

Step 4: Calculate moles H2SO4 need

For 2 moles Al we need 3 moles H2SO4

For 0.541 moles Al we need 3/2 * 0.541 = 0.8115 moles H2SO4

Step 5: Calculate mass H2SO4

Mass H2SO4 = 0.8115 moles * 98.08 g/mol

Mass H2SO4 = 79.6 grams

We need 79.6grams of H2SO4

Viefleur [7K]4 years ago
3 0

Answer:

76.3 g of H₂SO₄ are needed in this reaction

Explanation:

Let's begin with the reaction:

2Al(s) + 3H₂SO₄(aq)  →  Al₂(SO₄)₃ (aq)  +  3H₂(g)

Ratio in the reactants side is 2:3. It means that 2 moles of aluminum need 3 moles of sulfuric acid to react.

We determine the mass of Al → 14.6 g . 1mol / 26.98 g = 0.519 moles

Let's propose this rule of three:

2 moles of Al react with 3 moles of H₂SO₄

Then 0.519 moles of Al will react with (0.519 . 3) /2 = 0.778 moles of acid.

We convert the moles to mass, to find the answer:

0.778 mol . 98 g / 1mol = 76.3 g

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Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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3 years ago
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
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Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

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\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

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Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

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g If you have three identical containers (same volume) at the same temperature and pressure, each with a different gas. Containe
vekshin1

Answer:

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Explanation:

If these three gases (Helium He, Neon Ne, and Oxygen O_{2}) are all contained in separate identical containers with the same volume. And they are all stored at the same temperature, and pressure. Then, they'll all contain the same number of molecules. This is in line with Avogadro's law which states that "Equal volume of all gases, at the same temperature and pressure, have the same number of molecules."

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Solution:

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