Sodium azide, the explosive chemical used in atuomobile airbags, is made by the following reactions:

NaNO3 + 3NaNH2 => NaN3 + 3NaOH+ NH3

If you combine 15,0 g of NaNO3 (85 g/mol) with 15.0 g of NaNH3, what mass of NaN3 is produced?

You have the equation.

Convert 15.0 g NaNO3 to moles. #moles = grams/molar mass.

Convert moles NaNO3 to moles NaN3 using the coefficients in the balanced equation.

Convert moles NaN3 to grams by
grams = moles x molar mass

Post your work if you get stuck.

To determine the mass of NaN3 produced, we first need to find the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reagent, we can compare the amount of moles of NaNO3 and NaNH2. Let's calculate the number of moles of each reactant:

Moles of NaNO3 = mass / molar mass = 15.0 g / 85 g/mol = 0.1765 mol
Moles of NaNH2 = mass / molar mass = 15.0 g / 39.0 g/mol = 0.3846 mol

According to the balanced equation, the stoichiometric ratio between NaNO3 and NaNH2 is 1:3. This means that for every 1 mole of NaNO3, we need 3 moles of NaNH2.

From the given amounts of NaNO3 and NaNH2, we can see that NaNO3 is the limiting reagent because it has fewer moles compared to NaNH2. Therefore, we will use the moles of NaNO3 to calculate the moles of NaN3 formed.

Moles of NaN3 = moles of NaNO3 x (1 mol NaN3 / 1 mol NaNO3)
= 0.1765 mol x (1 mol NaN3 / 1 mol NaNO3)
= 0.1765 mol

Now, to find the mass of NaN3 formed, we can use the moles of NaN3 and its molar mass:

Mass of NaN3 = moles of NaN3 x molar mass of NaN3
= 0.1765 mol x 65 g/mol
= 11.46 g

Therefore, the mass of NaN3 produced is 11.46 grams.