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Sodium reacts with oxygen to produce sodium oxide. $$ 4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s) $$ a. How many grams of \(\mathrm{Na}_{2} \mathrm{O}\) are produced when \(57.5 \mathrm{~g}\) of Na reacts? b. If you have \(18.0 \mathrm{~g}\) of \(\mathrm{Na}\), how many grams of \(\mathrm{O}_{2}\) are needed for the reaction? c. How many grams of \(\mathrm{O}_{2}\) are needed in a reaction that produces \(75.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
a. 77.48 g \(\mathrm{Na}_{2}\mathrm{O}\), b. 6.27 g \(\mathrm{O}_{2}\), c. 19.36 g \(\mathrm{O}_{2}\)

Step by step solution

01

Write the balanced chemical equation

The given equation is already balanced: \[ 4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2}\mathrm{O}(s) \].
02

Calculate the molar masses

Find the molar masses of the substances involved: - Molar mass of Na: 22.99 g/mol - Molar mass of \(\mathrm{Na}_{2}\mathrm{O}\): 61.98 g/mol
03

Calculate moles of Na for part (a)

Use the molar mass to convert grams of Na to moles: \[ n = \frac{m}{M} = \frac{57.5 \text{ g}}{22.99 \text{ g/mol}} = 2.50 \text{ mol} \text{ of Na} \].
04

Stoichiometry for Na to \(\mathrm{Na}_{2}\mathrm{O}\)

From the balanced equation, 4 moles of Na produce 2 moles of \(\mathrm{Na}_{2}\mathrm{O}\). Set up the ratio: - \[ \text{Moles of } \mathrm{Na}_{2}\mathrm{O} = \frac{2.50 \text{ mol Na}}{4} \times 2= 1.25 \text{ mol } \mathrm{Na}_{2}\mathrm{O} \].
05

Convert moles of \(\mathrm{Na}_{2}\mathrm{O}\) to grams

Use the molar mass to find the mass of \(\mathrm{Na}_{2}\mathrm{O}\) produced: \[ m = n \times M = 1.25 \text{ mol } \times 61.98 \text{ g/mol} = 77.48 \text{ g } \mathrm{Na}_{2}\mathrm{O} \].
06

Calculate moles of Na for part (b)

For 18.0 g of Na, calculate the moles: \[ n = \frac{m}{M} = \frac{18.0 \text{ g}}{22.99 \text{ g/mol}} = 0.783 \text{ mol } \text{ of Na} \].
07

Stoichiometry for Na to \(\mathrm{O}_{2}\)

From the balanced equation, 4 moles of Na react with 1 mole of \(\mathrm{O}_{2}\). Set up the ratio: \[ \text{Moles of } \mathrm{O}_{2} = \frac{0.783 \text{ mol Na}}{4} = 0.196 \text{ mol } \mathrm{O}_{2} \].
08

Convert moles of \(\mathrm{O}_{2}\) to grams

Use the molar mass of \(\mathrm{O}_{2}\) (32.00 g/mol) to find the mass needed: \[ m = n \times M = 0.196 \text{ mol } \times 32.00 \text{ g/mol} = 6.27 \text{ g } \mathrm{O}_{2} \].
09

Calculate moles of \(\mathrm{Na}_{2}\mathrm{O}\) for part (c)

For 75.0 g of \(\mathrm{Na}_{2}\mathrm{O}\), calculate the moles: \[ n = \frac{m}{M} = \frac{75.0 \text{ g}}{61.98 \text{ g/mol}} = 1.21 \text{ mol } \mathrm{Na}_{2}\mathrm{O} \].
10

Stoichiometry for \(\mathrm{Na}_{2}\mathrm{O}\) to \(\mathrm{O}_{2}\)

From balanced equation, 2 moles of \(\mathrm{Na}_{2}\mathrm{O}\) come from 1 mole of \(\mathrm{O}_{2}\). Set up the ratio: \[ \text{Moles of } \mathrm{O}_{2} = \frac{1.21 \text{ mol }}{2} = 0.605 \text{ mol } \mathrm{O}_{2} \].
11

Convert moles of \(\mathrm{O}_{2}\) to grams

Use the molar mass of \(\mathrm{O}_{2}\) to find the mass needed: \[ m = n \times M = 0.605 \text{ mol } \times 32.00 \text{ g/mol} = 19.36 \text{ g } \mathrm{O}_{2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

chemical reactions
Chemical reactions involve the transformation of reactants into products. This is represented by a chemical equation. For the reaction between sodium (Na) and oxygen (\text{O}_{2}), the equation is:

\[ 4 \text{Na}(s) + \text{O}_{2}(g) \longrightarrow 2 \text{Na}_{2}\text{O}(s) \]

This equation shows that four sodium atoms react with one molecule of oxygen to produce two molecules of sodium oxide (\text{Na}_{2}\text{O}). Understanding the behavior of these substances on the atomic level helps in predicting the outcome of reactions. By following the law of conservation of mass, we ensure that all atoms present in the reactants are accounted for in the products.
mole calculations
Mole calculations are fundamental in stoichiometry as they allow us to quantify the amount of substances involved in a chemical reaction. The mole is a unit that represents a specific number of particles, typically atoms or molecules.

To find the number of moles, we use the formula: \[ n = \frac{m}{M} \]

Here, \(n\) is the number of moles, \(m\) is the mass of the substance, and \(M\) is the molar mass. For example, to find how many moles of sodium are in 57.5 grams:

\[ n = \frac{57.5 \text{ g}}{22.99 \text{ g/mol}} = 2.50 \text{ mol} \text{ of Na} \]

This calculation allows us to proceed with stoichiometry to determine how much product is formed or how much reactant is needed.
balanced equations
A balanced equation ensures that the same number of each type of atom is present on both sides of the reaction. This is essential because the law of conservation of mass states that matter cannot be created or destroyed.

For instance, in the sodium and oxygen reaction, the balanced equation is:

\[ 4 \text{Na}(s) + \text{O}_{2}(g) \longrightarrow 2 \text{Na}_{2}\text{O}(s) \]

It shows that four sodium atoms combine with one oxygen molecule to produce two sodium oxide units. Balancing involves adjusting coefficients (the numbers in front of the substances) to ensure that the same number of each atom appears on both sides of the equation. This is crucial for accurate stoichiometric calculations.
molar mass
Molar mass is the mass of one mole of a substance, which is useful in converting between mass and moles. It is expressed in grams per mole (g/mol).

For elements, it corresponds to the atomic mass listed on the periodic table. For compounds, it is the sum of the atomic masses of all the atoms in the molecule.

For example, the molar mass of sodium (\text{Na}) is 22.99 g/mol. For sodium oxide (\text{Na}_{2}\text{O}), the molar mass is calculated by adding the masses of two sodium atoms and one oxygen atom:

\[ \text{(2}\times 22.99 \text{ g/mol)} + 16.00 \text{ g/mol} = 61.98 \text{ g/mol} \]

Knowing the molar mass allows us to convert weights into moles, which is essential for stoichiometric calculations predicting how much product will form in a given reaction, or how much reactant is required.

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