Chemistry: Oxygen is produced by the reaction of sodium peroxide and water. 2na2o2(s) + 2h2o(l) o2(g) + 4naoh(aq) a. calculate the mass of na2o2 in grams needed to form 4.80g of oxygen. b. how many grams of naoh are produced when 4.80 g of o2 is formed? c.when 0.48g of na2o2 is dropped in water, how many grams of o2 are formed?

Oxygen is produced by the reaction of sodium peroxide

Ответ:

tayacall9362

07.01.2022

Chemistry

For a: The mass of Na_2O_2 required is 23.4 grams

For b: The mass of NaOH required is 24 grams

For c: The mass of oxygen gas produced is 0.0984 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(2)

For oxygen gas:

Given mass of oxygen gas = 4.80 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 2, we get:

$\text{Moles of oxygen gas}=\frac{4.80g}{32g/mol}=0.15mol$

For the given chemical equation:

$2Na_2O_2(s)+2H_2O(l)\rightarrow O_2(g)+4NaOH(aq.)$

For a:

By Stoichiometry of the reaction:

1 mole of oxygen gas is produced from 2 moles of Na_2O_2

So, 0.15 moles of oxygen gas will be produced from = $\frac{2}[1}\times 0.15=0.3mol$ of Na_2O_2

Now, calculating the mass of Na_2O_2 by using equation 1:

Molar mass of Na_2O_2 = 78 g/mol

Moles of Na_2O_2 = 0.3 moles

Putting values in equation 1, we get:

$0.3mol=\frac{\text{Mass of }Na_2O_2}{78g/mol}\\\\\text{Mass of }Na_2O_2=(0.3mol\times 78g/mol)=23.4g$

Hence, the mass of Na_2O_2 required is 23.4 grams

For b:

By Stoichiometry of the reaction:

When 1 mole of oxygen gas is produced, then 4 moles of NaOH is also produced.

So, when 0.15 moles of oxygen gas will be produced, then = $\frac{4}[1}\times 0.15=0.6mol$ of NaOH will be produced.

Now, calculating the mass of NaOH by using equation 1:

Molar mass of NaOH = 40 g/mol

Moles of NaOH = 0.6 moles

Putting values in equation 1, we get:

$0.6mol=\frac{\text{Mass of NaOH}}{40g/mol}\\\\\text{Mass of NaOH}=(0.6mol\times 40g/mol)=24g$

Hence, the mass of NaOH required is 24 grams

For c:

Given mass of Na_2O_2 = 0.48 g

Molar mass of Na_2O_2 = 78 g/mol

Putting values in equation 2, we get:

$\text{Moles of }Na_2O_2=\frac{0.48g}{78g/mol}=6.15\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of Na_2O_2 produces 1 mole of oxygen gas

So, $6.15\times 10^{-3}$ moles of Na_2O_2 will produce = $\frac{1}[2}\times 6.15\times 10^{-3}=3.075mol$ of oxygen gas

Now, calculating the mass of oxygen gas by using equation 1:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = $3.075\times 10^{-3}$ moles

Putting values in equation 1, we get:

$3.075\times 10^{-3}mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(3.075\times 10^{-3}mol\times 32g/mol)=0.0984g$

Hence, the mass of oxygen gas produced is 0.0984 grams.

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