given the following balanced chemical equation, how many mol of NaN3 can be form
ID: 603927 • Letter: G
Question
given the following balanced chemical equation, how many mol of NaN3 can be formed if 2.50 mol of Na are allowed to react with 4.50 mol of N2? 2Na +3N2 --> 2NaN3A. 2.50 mol NaN3
B. 3.00 mol NaN3
C. 4.50 mol NaN3
D. 7.00 mol NaN3
Explanation / Answer
How many liters of gas can be formed by heating 39.5 g of NaN3? If 1.00 mol of N_2 has a volume of 47.0 L under the reaction conditions, how many liters of gas can be formed by heating 39.5 g of NaN_3? The balanced equation: 2NaN3 --> 3N2 +2Na First, you need to convert 39.5g of NaN3 to # moles by using MW of NaN3 (MW =molecular weight) MW of NaN3 (from periodic table) = Na + 3x N = 23 + 3(14) =65g/mol meaning: 1 mol of NaN3 give 65g of NaN3 39.5 g of NaN3 x (1mol / 65g) =0.608 moles of NaN3 From the balanced equation: every 2 moles of NaN3 gives 3 moles of N2 0.608 moles of NaN3 x (3 moles N2/ 2moles NaN3) = 0.912 moles of N2 Given that 1.00mol of N2 has 47.0L then 0.912 moles of N2 x (47.0L/1.00mol) = 42.84230769 L = 42.8 L (3 significant figures)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.