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lidiya [134]
3 years ago
10

Read the given equation. Na2O2 + CO2 → Na2CO3 + O2 What volume of O2 gas is produced from 2.80 liters of CO2 at STP? 5.60 liters

4.20 liters 2.10 liters 1.40 liters
Chemistry
2 answers:
OleMash [197]3 years ago
8 0

The equation should be balanced as follows  2Na2O2 + 2CO2⇒ 2Na2CO3 +O2

2 moles of carbon (IV) oxide give 1 mole of oxygen.

Therefore, the mole ratio is of CO2 to O2 is 2:1

if 2.8 liters of CO2 were used, then the volume of O2 produced will be given by:   (2.80liters× 1)/2= 1.40 liters

umka21 [38]3 years ago
3 0

Answer:

The correct answer is 1.40 L.

Explanation:

Volume of CO_2 = 2.80 L

At STP. 1 mol of gas occupies = 22.4 L

Then 2.80 L of volume will be occupied by:

\frac{1}{22.4 L}\times 2.80 L=0.125 mol

2Na_2O_2 + 2CO_2 \rightarrow 2Na_2CO_3 + O_2

Acording to reaction 2 mol of CO_2 gives 1mol of O_2

Then 0.125 mol of CO_2 will give:

\frac{1}{2}\times 0.125 mol=0.0625 mol of O_2

Volume occupied by 0.0625 mol of oxygen gas at STP :

0.0625 \times 22.4 L= 1.40 L

1.40 L of oxygen gas will be produced.

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6 0
3 years ago
How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?
podryga [215]

Moles of Oxygen= 2.8075 moles

<h3>Further explanation</h3>

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

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7 0
3 years ago
What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?
serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0
3 years ago
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