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3 years ago

What quantities are conserved when balancing a chemical reaction?

1 answer:

8 0

The mass in a chemical reaction remains (mostly) the same.

(except for radiation/nuclear fission, in which mass gets converted into energy)

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Based on the given description which is most likely a mixture

suter [353]

Answer:

what descriptions

Explanation:

6 0

3 years ago

How many grams of aluminum sulfate: Al2(SO4)3 , would be formed if 250 g of H2SO4 react with aluminum? The reaction is: 2Al + 3H

NikAS [45]

Answer:

290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

Explanation:

Step 1: Data given

Mass of H2SO4 = 250 grams

Molar mass H2SO4 = 98.08 g/mol

Step 2: The balanced equation

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Step 3: Calculate moles H2SO4

Moles H2SO4 = mass H2SO4 / molar mass H2SO4

Moles H2SO4 = 250 grams / 98.08 g/mol

Moles H2SO5 = 2.55 moles

Step 4: Calculate moles Al2(SO4)3

For 2 moles Al we need 3 moles H2SO4 to produce 1 mol Al2(SO4)3 and 3 moles H2

For 2.55 moles H2SO4 we'll have 2.55/3 = 0.85 moles Al2(SO4)3

Step 5: Calculate mass Al2(SO4)3

Mass Al2(SO4)3 = moles Al2(SO4)3 * molar mass

Mass Al2(SO4)3 = 0.85 moles * 342.15 g/mol

Mass Al2(SO4)3 = 290.8 grams

290.8 grams of aluminium sulfate (Al2(SO4)3) will be produced.

7 0

4 years ago

A chemist adds of a zinc nitrate solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has a

Ray Of Light [21]

Answer:

5.3 × 10⁻³ kg

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.</em>

<em />

We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:

0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol

The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:

2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g

1 kilogram is equal to 1000 grams. Then,

5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg

8 0

3 years ago

What is the oxidation number of a pure element?

Snowcat [4.5K]

The oxidation number of a pure element is zero.

3 0

3 years ago

Catalytic converters made of palladium (pd) reduce automobile pollution by catalyzing the reaction between unburned hydrocarbons

Doss [256]

Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates.

3 0

4 years ago