Question

QUESTION 3 Automobile airbags use the decomposition of sodium azide, NaN3, to provide gas for rapid inflation: 2 NaN3(s) + 2 Na(s) + 3 N2(g). Using stoichiometry and the ideal gas law, calculate the mass (in g) of NaN3 required to provide 45.7 L of N2(g) at 41.4 °C and 1.00 atm? QUESTION 4 Consider the reaction between hydrogen gas and oxygen gas to form water: 2 H2(g) + O2(g) + 2 H2O(g). How many grams of water could be produced by the reaction of 7.51 liters of hydrogen with 4.11 liters of oxygen at STP? (Hint: check for limiting reagent).

answer both questions pls!!

Answer 1

step find moles of Na. we know, PVENRT na hatm X 45.7L 0.0821 atm L x(41.4 +273015) molok n= x 45.7 000821 8 314055 n= 1.75 mol Step 1 find mass of NaN3 2 Na Ng(8) 4 2 Na(s) + 3N268) applied POAC Rule. moles of Nang & moles of Na officient cofficient mass of Nana 1077 molar mass & 2 3 mass of Nal = 1.77 6582 3 mass of NaN3 = 76.68g

Aus given latim. at stl, T = 298 K P = step find moles of H₂ and O₂ for Hy ve 7.51 L PVE ART n= I x 7.51 0.307 mol 0.0821 x 298 for O2 z ve 4.11 n = PV - 1X4011 RT 0.0818 298 - 00168 mal Step D). find limiting Reagento 2H₂ (8) + O₂(g) → 2H2O(8) a) mole her officient of H2 = moles cofficient = 0.307 - 0.1535 b) mole per cofficient of O2 = 0.168 200/68 so, H2 Act as limiting sergent.

Step III find mass of the 2120 2 H₂(g) + q (g) applied PORC moles of H₂O & moles of H2 (officient Cofficient. moles of H20 - 0.307 mass of H2O & = moles & molan mas 0.307 X 18 = 5.52 g Thank you May it help you