8 - Stoichiometry

limiting reactant

The reactant that determines the amount of product formed in a chemical reaction the reactant that makes the least amount of product the reactant that is completely consumed in a chemical reaction.

percent yield

the percentage of the theoretical yield that was actually attained, would be: actual yield/theoretical yield x 100

Find the limiting reactant for each of the initial quantities of reactants. 2Na(s)+Br2(l)→2NaBr(s) 1.8 mol Na, 1 mol Br2

A. use conversion 2 mol Na makes 2 mol NaBr B. 1.8 mol Na x (2 mol NaBr / 2 mol Na) = 1.8 mol C. use conversion 1 mol Br2 makes 2 mol NaBr D. 1 mol Br2 x (2 mol NaBr / ! mol Na) = 2 mol E. the lesser one is the limiting reactant F. Na is the limiting reactant

Find the limiting reactant for each of the initial quantities of reactants. 2Na(s)+Br2(l)→2NaBr(s) 15.0 mol Na, 8.0 mol Br2

A. use conversion 2 mol Na makes 2 mol NaBr B. 15.0 mol Na x (2 mol NaBr / 2 mol Na) = 15 mol C. use conversion 1 mol Br2 makes 2 mol NaBr D. 8.0 mol Br2 x (2 mol NaBr / ! mol Na) = 16 mol E. the lesser one is the limiting reactant F. Na is the limiting reactant

Find the limiting reactant for each of the initial quantities of reactants. 2Na(s)+Br2(l)→2NaBr(s) 2.2 mol Na, 1 mol Br2

A. use conversion 2 mol Na makes 2 mol NaBr B. 2.2 mol Na x (2 mol NaBr / 2 mol Na) = 2.2 mol C. use conversion 1 mol Br2 makes 2 mol NaBr D. 1 mol Br2 x (2 mol NaBr / ! mol Na) = 2 mol E. the lesser one is the limiting reactant F. Br2 is the limiting reactant

Calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) 1.015x10^-3

A. use conversion if 4 mol NO2 make 2 mol N2O5 then how many mol NO2 make 1.5 mol N2O5 B. 1.015x10^-3 x (4 mol NO2 / 2 mol N2O5) = 2.03x10^-3 mol NO2

Calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) 1.5 mol N2O5

A. use conversion if 4 mol NO2 make 2 mol N2O5 then how many mol NO2 make 1.5 mol N2O5 B. 1.5 mol N2O5 x (4 mol NO2 / 2 mol N2O5) = 3 mol NO2

Calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) 5.4 mol N2O5

A. use conversion if 4 mol NO2 make 2 mol N2O5 then how many mol NO2 make 5.4 mol N2O5 B. 5.4 mol N2O5 x (4 mol NO2 / 2 mol N2O5) = 11 mol NO2

Calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) 4.87x10^3 mol N2O5

A. use conversion if 4 mol NO2 make 2 mol N2O5 then how many mol NO2 make mol N2O5 B. 4.87x10^3 mol N2O5 x (4 mol NO2 / 2 mol N2O5) = 9740 mol NO2

If the theoretical yield of a reaction is 23.9 g and the actual yield is 18.8 g , what is the percent yield?

18.8 g / 23.9 g = 78.7%

Calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2HgO(s)→2Hg(l)+O2(g) 1.49 kg HgO

A. need to know the molar mass of HgO (216.59) and O2 (32) B. convert kg to g so, 1.49 kg = 1490 g C. use conversion 1 mol O2 makes 2 mol HgO in middle D. 1490 g HgO x (1 mol HgO / 216.59 g HgO) x (1 mol O2 / 2 mol HgO) x (32 g O2 / 1 mol O2) = 110 g O2

Calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2HgO(s)→2Hg(l)+O2(g) 2.01 g HgO

A. need to know the molar mass of HgO (216.59) and O2 (32) B. use conversion 1 mol O2 makes 2 mol HgO in middle C. 2.01 g HgO x (1 mol HgO / 216.59 g HgO) x (1 mol O2 / 2 mol HgO) x (32 g O2 / 1 mol O2) = 0.148 g O2

Calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2HgO(s)→2Hg(l)+O2(g) 3.93 mg HgO

A. need to know the molar mass of HgO (216.59) and O2 (32) B. use conversion 1 mol O2 makes 2 mol HgO in middle C. convert mg to g so, 3.93 mg = 3.93 x 10^-3 D. 3.93 x 10^-3 g HgO x (1 mol HgO / 216.59 g HgO) x (1 mol O2 / 2 mol HgO) x (32 g O2 / 1 mol O2) = 2.90 x 10^-4 g O2

Calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2HgO(s)→2Hg(l)+O2(g) 6.85 g HgO

A. need to know the molar mass of HgO (216.59) and O2 (32) B. use conversion 1 mol O2 makes 2 mol HgO in middle C. 6.85 g HgO x (1 mol HgO / 216.59 g HgO) x (1 mol O2 / 2 mol HgO) x (32 g O2 / 1 mol O2) = 0.506 g O2

Find the limiting reactant for each of the initial quantities of reactants. 2Na(s)+Br2(l)→2NaBr(s) 1 mol Na , 1 mol Br2

A. use conversion 2 mol Na makes 2 mol NaBr B. 1 mol Na x (2 mol NaBr / 2 mol Na) = 1 mol C. use conversion 1 mol Br2 makes 2 mol NaBr D. 1 mol Br2 x (2 mol NaBr / ! mol Na) = 2 mol E. the lesser one is the limiting reactant F. Na is the limiting reactant