How many grams of nan3 are required to produce 10 l of nitrogen gas if the gas has a density of 1.25 g/l?

Given information : Volume of nitrogen gas N2 =10 L

Density of nitrogen gas N2 = 1.25 g/L

We need to find grams of NaN3.

To find the grams of NaN3 we first need to find the grams of N2 and then using stoichiometry we can calculate grams of NaN3 using grams of N2.

Step 1 : Find the mass (grams) of N2 using its volume and density.

Mass (grams) = Density X Volume

Mass of N2 = 12.5 g

Step 2 : Convert grams of N2 to moles of N2

Gram of N2 = 12.5 g and Molar mass of N2 = 28g/mol

Moles of N2 = 0.446 mol

Step 3 : Calculate moles of NaN3 using moles of N2 and molar ratio.

To find the moles of NaN3 we first need to write the balanced chemical equation.

A molar ratio is the ratio between the amounts in moles of any two compounds involved in a balanced chemical reaction.It is determined by examining the coefficients in front of formulas in a balanced chemical equation. Coefficients are the number present in front of the compound or formula.

Coefficient of NaN3 is 2 and coefficient of N2 is 3, so mole ratio of NaN3 to N2 is 2:3

Mole of NaN3 =

Mole of NaN3 = 0.297 mol NaN3

Step 4 : Convert moles of NaN3 to grams of NaN3

Grams = Molar mass X moles

Molar mass of NaN3 = 65 g/mol and Moles = 0.297 mol NaN3

Grams of NaN3 = 19.3 gram(g)

19.3 grams of NaN3 is required  to produce 10 L of nitrogen gas.

the answer is barium and copper bromide