I once again am stuck on some problems in this chapter. While there is many more I don't get these six I think would help me with the others I'm stuck on. Would highly appreciate very detailed work shown so I can understand the process and apply to other questions of like nature.
1. Using the the following equation: 2 NAOH + H2SO4 --> 2 H2O + Na2SO4
How many grams of Sodium Sulfate will be formed if you start with 200g of sodium hydroxide and you have an excess of sulfuric acid?
2. Using the following equation: Pb(SO4)2 + 4 LiNO3 --> Pb(NO3)4 + 2 LiSO4
How many grams of lithium nitrate will be needed to make 250g of lithium sulfate, assuming that you have an adequate amount of Lead (IV) sulfate to do the reaction?
3a. In the reaction 2 C8H18 + 25 )2 --> 16 CO2 + 18 H2O, what is the ratio of volumes of O2 to CO2?
3b. If 27.3g of C8H18 are combusted, what mass of water will be produced?
3c. How many molecules of CO2 will be produced?
4. F2 (g) + NH3 (g) --> N2F2 (g) + HF (g)
Identify the limiting reagent for the given combination of reactants. Calculate how many grams of whatever substance will be in excess and how many grams of HF will be made.
5. In a laborator, Qem disolved 11.2 g of sugar in water, and then he poured the solution into a 250-mL volume flask. He added enough water to make the solution exactly 250.00mL. What is the concentration of the solution he prepared?
6. CH4 + 2 O2 --> 2 H2O + CO2
a) If 50.0 L of methane at STP are burned, what volume of carbon dioxide will be produced at STP?
b) If 50.0 L of methane at RTP are burned, what volume of gaseous water at STP is produced?
*If someone could tell me what STP and RTP are that would also be nice*
Please see the attachment.
Question 4 is finished here.
4. Balance the equation, 4F2 + 2NH3 = N2F2 + ...
The expert determines how many grams of Sodium Sulfate will be formed if you start with 200g of sodium hydroxide.
Stoichiometry in Reactions of Aluminum and Sodium Hydroxide
6) Explain why the reaction of 31.1636 g of aluminum metal (atomic mass 26.9815 g/mol) with 110.00 g of sodium iodate (molar mass 197.89 g/mol) in the presence of nitric acid produces the equivalence of only 25.00 g of aluminum ions; whereas,reaction with 120.00 g of sodium iodate produces the equivalence of 31.1636 g of aluminum ions.
7) You place 129.50 g of sodium hydroxide into exactly 1635 mL of water. With how many mL of 1.9801 M hydrochloric acid need to be added to react completely (i.e.,titrate) the sodium hydroxide? What will the final concentration of the sodium chloride be in the mixture?View Full Posting Details