University:
University of North Carolina at Charlotte
Course:
CHEM 1251 | General Chemistry I
Academic year:

2024
Views:

325

Pages:

7
Author:

Kamila P.

Practice Show how in photosynthesis, plants convert gaseous carbon dioxide and liquid water into aqueous glucose (C6H12O6) and gaseous oxygen. Write the balanced reaction below. 6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) A particular plant consumes 37.8 g of CO2 in one week. Assuming that there is more than enough water present to react with all of CO2, what mass of glucose (in grams) can the plant synthesize in one week? 37.8 g CO2 × 1 mol CO2 × 1 mol C6H12O6 × 180.156 g C6H12O6 / 44.01 g CO2 × 6 mol CO2 × 1 mol C6H12O6 = 25.89 g C6H12O6 Practice Mass-to-Mass Conversion Determine the mass (in grams) of CO2 produced when 3.6 x 10^15 g of C6H18 is burned in excess oxygen gas. 2 C6H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l) 3.6 x 10^15 g of C6H18 × 1 mol of C6H18 × 16 moles of CO2 × 44.01 g CO2 / 114.224 g C6H18 × 2 moles of C6H18 × 1 mol CO2 = 1.1 x 10^16 g CO2 EXAM 3 PRACTICE H-N-H H d+ -N-d- H [O-N=O]- [O=N-O]- Bent (MG), Trigonal Planar (EG) 3 electron domains IT'S POLAR BECAUSE OF THE LONE PAIR IFA: 3.5mol A, 5.9mol B, 2.2mol C 3A + 2B → C + 2D Moles of D from A = (3.5mol A) × (2mol D / 3mol A) = 2.33 mol D Moles of D from B = (5.9mol B) × (2mol D / 2mol B) = 5.9 mol D Moles of D from C = (2.2mol C) × (2mol D / 1mol C) = 4.4 mol D Excess B = 5.9 mol B - 2.3 mol B = 3.6 mol B ALEKS #10 Aqueous iron (III) chloride with solid magnesium to form solid iron and aqueous magnesium chloride. 2FeCl3(aq) + 3Mg(s) → 2Fe(s) + 3MgCl2(aq) Chlorine gas and solid sodium chloride are produced by the decomposition of solid sodium chloride. 4NaCl(s) → 2Cl2(g) + 2Na(l) 2NaCl(s) → Cl2(g) + 2Na(l) Dinitrogen pentoxide gas is produced by the reaction of nitrogen trioxide and nitrogen dioxide gas. why don't we put N2? NO3(g) + NO2(g) → N2O5(g) Solid iron reacts with aqueous copper (II) sulfate (CuSO4) to form aqueous iron (II) sulfate (FeSO4) and solid copper. Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s) Solid calcium hydroxide decomposes into gaseous water and solid calcium oxide. Ca(OH)2(s) → H2O(g) + CaO(s) H2O(g) + CaO(s) → Ca(OH)2(s) 2NH4ClO4(s) → N2(g) + Cl2(g) + 2O2(g) + 4H2O(g) moles NH4ClO4 consumed = 2 (1.6 mol) = 3.200... mol → 3.2 mol 4C2H2(g) + 5O2(g) → 8CO2(g) + 4H2O(g) moles CO2 consumed = 8/4 (0.10 mol) = 0.20 mol CH3CH2OH + O2(g) → H2O + CH3COOH moles CH3CH2OH = 4/4 (0.080 mol) = 0.020 mol H3PO4(aq) + 3NH3(g) → (NH4)3PO4(aq) moles H3PO4 = 1/1 (0.60 mol) = 0.60 mol Predicting Products from a General Statement: Stoichiometry (NH4)2CO3(aq) + 2NaOH(aq) → Na2CO3(aq) + 2H2O(l) + 2NH3(g) MnS(s) + 2HCl(aq) → H2S(g) + MnCl2(aq) Li2O(s) + 2HBr(aq) → 2LiBr(aq) + H2O(l) 2H2S(s) + 3O2(g) → 2MnO(s) + 2SO2(g) Co(s) + Cl2(g) → CoCl2(s) NaH(s) + H2O(l) → NaOH(aq) + H2(g) 2NaH(s) + 2H2O(l) → 2NaOH(aq) + H2(g) CH3CH2OH + O2(g) → H2O + CH3COOH ? mass of ethanol with 4.1 g of oxygen gas 4.1 g O2 × 1 mol O2 × 1 mol CH3CH2OH × 46.068 g CH3CH2OH / 32.00 g O2 × 1 mol O2 × 1 mol CH3CH2OH = 5.9 g CH3CH2OH 4NH3 + 5O2 → 4NO + 6H2O moles of oxygen gas consumed by the reaction: 8.6 g NH3 8.6 g NH3 × 1 mol NH3 × 5 mol O2 / 17.031 g NH3 × 4 mol NH3 × 1 mol O2 = 20 g O2