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VLD [36.1K]
2 years ago
8

A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t

he compound?
Chemistry
1 answer:
Effectus [21]2 years ago
5 0

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is Na_2CO_3 .

Hence, this is the required solution.

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Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and
alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

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Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

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What inspired Mendeleev to create his organization for the elements
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write the balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin(iv) acetate. if
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The reaction of Sodium hydroxide and Tin(iv) acetate is as follows:

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What is an Aqueous solution?

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5 0
10 months ago
Determine the empirical and molecular formula:
Bingel [31]

Answer:

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Explanation:

in 100% compound we have 75.6 % Carbon ( Molar mass = 12g/mole), 8.80% hydrogen ( Molar mass = 1.01 g/mole) and 15.5% Oxygen (Molar mass = 16.01 g/mole).

Carbon: 75.6g / 12 = 6.29

Hydrogen: 8.80/ 1 = 8.80

Oxygen: 15.5/ 16 = 0.97

⇒0.97 is the smallest so we divide everything through by 0.97

C: 6.29 / 0.97 =  6.48 ≈ 6.5

H: 8.80 /0.97 = 9

O: 0.97 / 0.97 = 1

To get rid of decimals, we multiply by 2  

C: 6.5 x 2 = 13

H: 9 x 2 = 18

O: 1 x 2 = 2

The empirical formula = C13H18O2

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