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1. One mole of an ideal gas occupies 22.4 L at STP (a) Calculate the mass of 11.2 L of oxygen gas at STP. (b) Calculate the number of atoms present in the above sample. 2. 21 g of nitrogen gas is mixed with 5 g of hydrogen gas to yield ammonia according to the equation. N2 + 3H2 → 2NH3 Calculate the maximum amount of ammonia that can be formed. |
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Answer» 1. (a) Mass of 22.4 L oxygen at STP = 32 g ∴ Mass of 11.2 L oxygen at STP = 16 g (b) No. of atoms present in 16 g of O2 \(\frac{6.02\times10^{23}}{2}\) ×2 = 6.02 × 1023 atoms 2. N2 + 3H2 → 2NH3 1 mole N2 + 3 mole H2 → 2moles NH3 1 mole N2 requires 3 mol H2 i.e., 28g N2 requires 6 g H2 Hence, 21 g N2 requires \(\frac{6\times21}{28}\) = 4.5 g H2 21g N2 reacts completely and 0.5g H2 remains unreacted. Hence, N2 is the limiting reagent. 28g N2 gives 2 x 17g NH3 ∴ 21 g N2 gives \(\frac{2\times17\times21}{28}\) = 25.5 g NH3 |
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