Answer:
The answer is 139.1 g of sodium azide.
Explanation:
To solve this question we need to find first the stoichometry of the reaction to find the moles involved.
1. The balanced formula:
2NaN3 ⇒ 3N2 + 2Na
(important: the nitrogen is a gas, therefore, is always diatomic N2)
2. The moles:
We need to convert the 90 g of N2 into moles to know the relations of amounts between the other compounds in the chemical equation
90 g of N2 x (1 mol of N2/ 28 g of N2) = 3.21 mol of N2
3. Stoichometry:
The relations of the amounts of moles of N2 and NaN3 using the balanced formula
3.21 mol of N2 x (2 mol of NaN3/3 mol of N2) = 2.14 mol of NaN3
4. Moles into grams
then to answer the question we convert the moles value of NaN3 into grams of NaN3
2.14 mol of NaN3 x ( 65 g of NaN3/ 1 mol of NaN3) = 139.1 g of NaN3.
and this is the answer of the question.
