PROBLEM 3.42Potassium Nitrate decomposes on heating, producing
potassium oxide and gaseous nitrogen and oxygen. To produce 56.6 Kg
of O2, how many a.) Moles of KNO3 and b.) Grams of KNO3 must be
heated?MODIFIED PROBLEMPotassium Nitrate decomposes on heating,
producing potassium oxide and gaseous nitrogen and oxygen. To
produce 100 Kg of O2, how many a.) Moles of KNO3 and b.) Grams of
KNO3 must be heated?SOLUTIONa.) 100 kg O2 X ((( = 25,000 mole
KNO3b.) (2,500 mole KNO3) X ( = 252,775 g KNO3
Problem 3.45Calculate the mass (g) of each product formed when
174 g of silver sulphide reacts with excess hydrochloric acid.Ag2S
(s) + HCl (aq) AgCl (s) + H2S (g) [unbalanced]MODIFIED
PROBLEMCalculate the mass (g) of each product formed when 200 g of
silver sulphide reacts with excess hydrochloric acid.Ag2S (s) + HCl
(aq) AgCl (s) + H2S (g) [unbalanced]SOLUTIONAg2S (s) + 2HCl (aq)
2AgCl (s) + H2S (g) [balanced]AgCl Mass = (200g Ag2Cl) ((( Ans:
231.383 g AgClH2S Mass = (200g Ag2Cl) ((( Ans: 25.082 g H2S
PROBLEM 3.60When 20.5 g of methane and 45.0 g of chlorine gas
undergo a reaction that has a 70.5% yield, what mass (g) of
chloromethane forms? Hydrogen Chloride also forms.MODIFIED
PROBLEMWhen 30 g of methane and 50 g of chlorine gas undergo a
reaction that has an 80% yield, what mass (g) of chloromethane
forms? Hydrogen Chloride also forms.SOLUTIONCH4(g) + Cl2(g)CH3Cl +
HCl(30 g CH4l) (( = 1.8703 mole CH3Cl(50 g Cl2) (( = 0.7052 mole
CH3Cl- Cl2 is the limiting reactantCH3Cl Mass = (0.7052 moles
CH3Cl) ( = 35.59 g CH3Cl(35.59 g CH3Cl) x 80% = 28.478 g CH3Cl
PROBLEM 3.64Butane Gas is compressed and used as a liquid fuel
in disposable cigarette lighters and lightweight camping stoves.
Suppose a lighter contains 5.50 mL of Butane (d = 0.579g/ml)a.) How
many grams of oxygen are needed to burn the butane completely?b.)
How many moles of H2O form when all the butane burns?c.) How many
total molecules of gas form when the butane burns
completely?MODIFIED PROBLEMSuppose a lighter contains 5.50 mL of
Butane (d = 0.579g/ml)a.) How many grams of oxygen are needed to
burn the butane completely?b.) How many moles of H2O form when all
the butane burns?c.) How many total molecules of gas form when the
butane burns completely?SOLUTION2C4H10(g) + 13O2(g) 8CO2(g) + H2O
(g)a.) (6.00 mL) ((( ( Ans: 12.45 g O2b.) (6.00 mL) (((Ans: 0.299
moles H2O c.) CO2 Moles = (6.00 mL)((( = 0.239 moles CO2Total Moles
= 0.299 moles H2O + 0.239 moles CO2
PROBLEM 3.75How many grams of NaH2PO4 are needed to react with
43.74 mL of 0.285 M NaOH.MODIFIED PROBLEMHow many grams of NaH2PO4
are needed to react with 50 mL of 0.300 M NaOH.SOLUTIONNaH2PO4 +
2NaOH Na3PO4 + 2H2ONaOH Moles = (50 mL x ( = 0.015 moles
NaOHNaH2PO4 Mass = (0.015 moles NaOH) (( = 0.899 g NaH2PO4
