Answers

1. If you had 5.6 g of CaO & unlimited H2O, how many grams of Ca(OH)2 could you get?

. CaO + H2O  Ca(OH )2Given: 5.6 g CaO ?: g Ca(OH)2

1) Make sure chemical equation is balanced correctly. {It is.}

2) Find molar mass of CaO from periodic table of elements. Ca = 40 g/m + O = 16 g/m = 56 g/m

3) Find # moles of CaO by dividing its # grams by its molar mass. 5.6 grams/ 56 g/m = 0.10 mol CaO

4) Write mole ratio from coefficients of the balanced equation, such that the # moles of the desired (sometimes called the “unknown”) is over the # moles of the given that you have just calculated.

[1 mol Ca(OH)2/ 1 mol CaO]

5) Multiply mole ratio by # moles just calculated to find the # moles of the unknown.

[1 mol Ca(OH)2/ 1 mol CaO] [0.10 mol CaO] = 0.10 mol Ca(OH)2

6) Multiply this # moles by the reagent’s molar mass (periodic table) to find # grams of the unknown.

Molar mass of Ca(OH)2 = 40 g/m + (2)(16g/m + 1 g/m) = 74 g/m

[0.10 mol Ca(OH)2] [74 g/m] = 7.4 gram Ca(OH)2

2. How many moles of LiOH is needed to make 5 moles of Fe(OH)3 if

Fe(NO3)3 + 3 LiOH  3 LiNO3 + Fe(OH)3

1) Equation is balanced.

2) Given: 5 mol Fe(OH)3Desired/unknown/ ? : mol LiOH

3) Mole ratio: (desired/ given): [3 mol LiOH/ 1 mol Fe(OH)3]

4) Find # mol unknown: (desired/ given)(actual) = wanted

[3 mol LiOH/ 1 mol Fe(OH)3] [5 mol Fe(OH)3] = 15 mol LiOH

3. One kilogram of Cu with 22.4 moles of nitric acid can make how many moles of NO?

Cu + HNO3 Cu(NO3)2 + H2O + NO(Must balance equation 1st)

1) Balance equation:3 Cu + 8 HNO3 3 Cu(NO3)2 + 4 H2O + 2 NO

2) Convert grams to moles for Cu: 1 kg = 1000 grams: 1000 grams/ 63.5 g/m = 15.75 mol Cu

3) Determine which reactant is the limiting reagent by using mole ratios and given # moles.

(3 mol Cu/ 8 mol HNO3)(22.4 mol HNO3) = 8.4 mol Cu needed to finish reaction, which you have.

Nitric acid is, therefore, the limiting reagent and its values are used in all subsequent reactions.

4) Use mole ratio of NO over HNO3 & multiply this by actual # mol of limiting reagent.

(2 mol N2 / 8 mol HNO3)( 22.4 mol HNO3) = 5.6 mol N2

4. You can produce how many grams of water by burning 54 L of acetylene (C2H2) in the presence of

166 L of oxygen?2 C2 H2 + 5 O24 CO2 + 2 H2O

1) Equation is balanced.

2) Convert Liters of gas to moles by dividing Liters by the gas constant 22.4 L/ mol.

166 L/ 22.4 L/mol = 7.4 L O2 & 54 L/ 22.4 L/ mol = 2.4 mol C2H2

3) Determine limiting reagent: (2 mol C2H2/ 5 mol O2)( 7.4 L O2) = 2.96 mol C2H2needed limited

(5 mol O2 / 2 mol C2H2)(2.4 mol C2H2) = 6 mol O2 needed & have 7.4 L

4) Use limiting reagent’s moles & mole ratio to find # moles for desired reagent (water).

(2 mol H2O/ 2 mol C2H2) (2.4 mol C2H2) = 2.4 mol H2O produced

5) Multiply # mol by its molar mass to find its mass in grams.

(2.4 mol H2O)(18 g/mol) = 43.2 grams H2O