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earnstyle [38]
1 year ago
14

Find the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius.

Chemistry
1 answer:
Rudiy271 year ago
8 0

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

n(Al)=0.6/27=0.02222 moles

n(O2)=n(Al)*3/4=0.016666 moles

m=n*M

m(O2)=0.0166666*32=0.53333 grams

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Answer:

The easiest way to identify a double displacement reaction is to check to see whether or not the cations exchanged anions with each other.

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3 years ago
Unit: Stoichiometry
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Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

7 0
3 years ago
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Answer: I think the answer is A

Explanation:

I’m sorry if I’m wrong.

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