# 3 3 3 1703 g nh 142755 mol nh 1 mol nh 24311 g nh 3

• Notes
• MCQubed
• 55
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 36 - 38 out of 55 pages.

3 3 3 17.03 g NH 1.42755 mol NH 1 mol NH = 24.311 g NH 3 % yield = actual yield 100 theoretical yield = 15.0 g 100 24.311 g = 61.700 = 61.7% 2 resent = 2 2 2 1 mol H 10.0 g H 2.016 g H = 4.960 2 b) Moles of H initially p 3 mol H Moles of N 2 initially present = 2 1 mol N 2 20.0 g N 28.02 g N = 0.71378 mol N 2 2 Moles of H 2 re red to oduce 15.0 g NH qui pr 3 = 3 2 3 15.0 g NH 17.03 g 3 3 H 3 mol H NH 2 mol NH   = 1.3212 mol H 2 1 mol N 3 2 3 3 3 1 mol NH 1 mol N  Moles of N 2 required to produce 15.0 g NH 3 = 15.0 g NH 17.03 g NH 2 mol NH   = 0.4404 mol N 2 Moles of H at equilibrium = initial moles – reacted moles = 4.9603 mol – 1.3212 mol = 3.6391 = 3.64 mol H 2 rium = initial moles – reacted moles = 0.71378 mol – 0 = 0.27338 = 0.273 mol N 2 2 Moles of N 2 at equilib .4404 mol 4-36
4.110 Plan: Ferrous ion is Fe 2+ . Write a reaction to show the conversion of Fe to Fe 2+ . Convert the mass of Fe in a 125-g Fe in a 737-g sample. Use molar mass f Fe to moles of Fe and use Avogadro’s number to convert moles of Fe to moles of ions. Solution: serving to the mass of to convert mass o = a) Fe oxidizes to Fe 2+ with a loss of 2 electrons. The H + in the acidic food is reduced to H 2 with a gain of 2 electrons. The balanced reaction is: Fe( s ) + 2H + ( aq ) Fe 2+ ( aq ) + H 2 ( g ) O.N.: 0 +1 +2 0 b) Mass (g) of Fe in the jar of tomato sauce 3 49 mg Fe 10 g 737 g sauce 125 g sauce 1 mg = 0.288904 g Fe ons = 2 23 2 2 1 mol Fe 1 mol Fe 6.022x10 mol Fe 1 mol   Number of Fe 2+ i Fe ions 0.288904 g Fe 55.85 g Fe 1 Fe   = 3.11509x10 21 = 3.1x10 21 Fe 2+ ions per jar of sauce 4.111 Plan: Write balanced equations for the conversion of CaCO 3 to CaO, the reaction of CaO with SO 2, and the three reactions to obtain the overall reaction. Use the mass % of d the mass and then moles of sulfur present. Use the molar ratio in the overall ass of CaCO 3 required to ct with this amount of sulfur, as ming a 70% yield. combustion of sulfur to produce SO 2 . Add these sulfur in the coal sample to fin balanced reaction to find the m rea su Solution: The reactions are: CaCO ( s )  CaO( 3 s ) + CO 2 ( g ) CaO( s ) + SO 2 ( g )  CaSO 3 ( s ) S( s ) + O 2 ( g )  SO 2 ( g ) Overall reaction: CaCO 3 ( s ) + S( s ) + O 2 ( g ) CaSO 3 ( s ) + CO 2 ( g ) Moles of sulfur = 3 4 10 g 0.33% 1 mol S 8.5x10 kg coal 1 kg 100% 32.07 g S = 8746.492 mol S Mass (g) CaCO 3 = 3 1 mol CaCO l S 3 3 100.09 g CaCO 100% 1 mol S 1 mol CaCO 70%  6 = 1.3x10 6 g CaCO 3 needed 4.112 Plan: 8746.492 mo = 1.25062x10 Convert the mass of glucose to moles and use the molar ratios from the balanced equation to find the and CO 2 . The amount of ethanol is co grams using its molar mass.
• • • 