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3 years ago

If 2.03 g of oxygen react with carbon monoxide, how many grams of CO2 are formed?

Chemistry

1 answer:

mamaluj [8]3 years ago

8 0

Answer:

The suitable equation for this reaction is

2CO + O₂ -----> 2CO₂

Here, we are given that we have 2 grams of O₂

From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂

Moles of O₂:

2/32 = 1/16 moles

Therefore, the number of moles of CO₂ is twice the moles of O₂

Moles of CO₂ = 2 * 1/16

Moles of CO₂ formed = 1/8 moles

Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂

Mass of CO₂ formed = 44 * 1/8

Mass of CO₂ formed = 5.5 grams

Hence, option B is correct

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3 years ago

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3 years ago

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If 0.600mol of chloride gas reacted with 0.500mol of aluminium metal to produce aluminium chloride,which reactant is in excess?h

Rus_ich [418]

Answer:

The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3

Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3

Given: 0.600 moleCl2; 0.500 mole Al

Required: Excess reactant___; Number of moles of AlCl3 produced__

Solution: Use dimensional analysis using the mole-mole rel

0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al

0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2

Based on the given:

0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)

0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)

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3 years ago

Calculate the percent composition of the following: A. VO3 B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

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3 years ago