Question

Please help with these two questions

1.) 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 4.28 liters of hydrogen with 4.11 liters of oxygen at STP?

2.) 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 42.4 L of N₂(g) at 44.1 °C and 1.00 atm?

QUESTION 3 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 4.28 liters of hydrogen with 4.11 liters of oxygen at STP? (Hint: check for limiting reagent). QUESTION 4 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 42.4 L of N2(g) at 44.1 °C and 1.00 atm? 1.1. a la. Lil . J L . 1. a. AU A .. 11

Answer 1

Groen condition is STP. ie. T=298k P = laty R = 0.08206 atm L mott ist. for H, V=4.28L. we have PV = nRT n = PV = 1x4. 28 RT (0.08206Y298) I n = 0.175 moles / for O2, V=4.11. in pu = 1 x Lill (0.08206)(298) n = 0.168 moles Givery Equation is 2H2 + O₂ 52,9) 0 ) for Every I mole of o. needed x 2 moles of the for 0.168 miles 10 - > 2x0.68 = 0.336 notes but were we have only 0.175 moles of the i lle is limiting seabent : Mow, a moles of Haco Gives 2 moles of talg of the Gives : 0.175 mol (²) xo . 1.75 mols of the 0.175 moles of H2c9) tive 0.17s moles of Hala, - Give

thank your you vmuch....

Now, & 0.175 mole of H₂O = 18 g. mole of H₂O= 18 %0.175 = 3.15 g Amount of wale produced - 3.15 g 2 Na N3G) > 2 Naic + 3 Na ig) Givens, T = 44.1°C= 273+44.1=317-1K V= 42.4L P= 1.00 atm. R = 0.08 206. Now, PU= nRT' n = PV = 1, 42-4 RT (0.08206)(319.1) In = 1.629 moles I from above Equation needs & 2 moles of Nanas) Mag) - 3 moles i mole of N2 (9) needs 2 moles of Na Asis) .. 1.629 moles of Na (9) - neods 2 x 1.629) moles of Nana : 1.629 moles of Me (g) need 1.086 moles of Nanse Now, 1 mole of AlaN3 - 5.00g of NaNa 1.086 moles of Manly 65.00 x 1.080= 70.60g. Amount of Nany needed = 40.60 g