How many moles of sodium nitrate, nano3, do you need to make 22.4l of oxygen gas at stp? *also if you can with the other questions i’d appreciate it

We need 2 moles of sodium nitrate NaNO3 fro the production of one mole of oxygen gas.

The balanced chemical equation is:

                   2NaNO3 →→→→ 2NaNO2 + O2.

From the balanced chemical equation, it is obvious that for the production of One molecule of oxygen two molecules of NaNO3 breakdown.

So 22.4 L is equal to one mole.

Hence for the production of one mole of oxygen gas two moles of sodium nitrate will be needed.

The grams of NaNO3 for the production of 23.98 L of oxygen gas is 181.8786.

From the balanced chemical equation is:

                   2NaNO3 →→→→ 2NaNO2 + O2.

For the production of one mole of oxygen we need two moles of NaNO3.

So for 22. 4 L of O2 = 2 moles of NaNO3.

For:

          1 L = 2/22.4

  23.98 L = 2/22.4 * 23.98 = 2.14 moles.

In one mole of NaNO3, there are 84.99 grams.

So in 2.14 moles:

Mass in grams = 2.14 * 84.99 = 181.8786 g

Hence the grams of NaNO3 for the production of 23.98 L of oxygen gas is 181.8786.

If 48.8 L of the oxygen gas is used in the reaction then 48.8 liters of the carbon mono oxide gas will produce.

From the balanced chemical equation:

                C6H6S + 6O2 →→→→  6CO + 3H2O + SO3

In the production of the six carbon monoxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

So for the production of 48.8 L of carbon mono oxide production, the 48.8 L of the oxygen gas will be needed.

The 0.005648 L of the CO will produce with the use of 0.005648 L of oxygen gas.

From the balanced chemical equation:

                C6H6S + 6O2 →→→→  6CO + 3H2O + SO3

In the production of the six carbon mono oxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

One mole of any gas is equal to 22.4 L.

So for the production of 0.005648 L of carbon mono oxide production, the  0.005648 L of the oxygen gas will be needed.

The 2.98 L of the carbon monoxide will be produced if we use 2.98 liters of oxygen gas.

From the balanced chemical equation, we can predict the amount of reactants and products in the chemical equation.

From the balanced chemical equation:

                C6H6S + 6O2 →→→→  6CO + 3H2O + SO3

In the production of the six carbon mono oxide six oxygen molecules are used up.

It means the ratio is 1 : 1.

It means the the amount of CO produced will be equal to the amount of O2 used.

One mole of any gas is equal to 22.4 L.

Hence for use of 2.97 L of oxygen gas, the 2.98 L of the carbon monoxide will be produced

there are two significant figures.