How many grams of sodium azide are required to produce 31 grams of nitrogen
balance the dang reaction equation!
2NaN3>> 3N2 + 2Na
so it takes 2/3 moles as moles of N2.
gramsNaN3=2/3 * 31/formulmassN2 * molmassNaN3
To determine how many grams of sodium azide are required to produce 31 grams of nitrogen, we first need to know the chemical equation for the reaction in which sodium azide produces nitrogen.
The chemical equation for the reaction is:
2 NaN3 → 2 Na + 3 N2(g)
From the equation, we can see that 2 moles of sodium azide (NaN3) produce 3 moles of nitrogen (N2). To convert moles to grams, we need to know the molar mass of each compound.
The molar mass of sodium azide (NaN3) is calculated as follows:
Na = 22.99 g/mol
N = 14.01 g/mol
Multiply the molar mass of nitrogen (N) by 3 since there are 3 moles of nitrogen in the reaction:
3 × 14.01 g/mol = 42.03 g/mol
Now, to find out how many grams of sodium azide are required, we can set up a proportion using the molar mass and the given mass of nitrogen:
2 moles of NaN3 / 3 moles of N2 = x grams of NaN3 / 31 grams of N2
Solving the proportion:
2/3 = x/31
Cross-multiply:
2 × 31 = 3x
62 = 3x
Divide by 3:
x = 62/3
x ≈ 20.67 grams
Therefore, approximately 20.67 grams of sodium azide are required to produce 31 grams of nitrogen.