Calculate the mass of sodium nitrate( NaNO3), in grams, required to produce 128 g of oxygen, using the following equation: 2NaNO3 --> 2NaNO2 O2

680g of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

2NaNO3 —> 2NaNO2 + O2

Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol

Mass of NaNO3 from the balanced equation = 2 x 85 = 170g

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32g

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:

From the balanced equation above,

170g of NaNO3 decomposed to produce 32g of O2.

Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e

Xg of NaNO3 = (170 x 128)/32

Xg of NaNO3 = 680g

Therefore, 680g of NaNO3 are needed to produce 128g of O2.

oh it is true if thats what you are asking

if you mean hydrogen and oxygen then water is the product