find the mass of oxygen to react with 176g of propane
Answers
Answer:
Explanation:1
U5LM2B-WS - Stoichiometric Relations Name: _____KEY_____
Consider the following chemical equation to answer the questions below.
C3H8 + 5O2 3CO2 + 4H2O
1. _One_ molecules of C3H8 react with _five_ molecules of O2 to produce _three_ molecules of
CO2 and _four_ molecules of H2O. Coefficients in the balanced chemical equation can be
read in terms of molecules or moles.
2. 20 molecules of C3H8 react with _100_ molecules of O2 to produce _60_ molecules of CO2 and
_80_ molecules of H2O. Use stoichiometric ratios from the balanced equation.
Ex: 20 molecules of C3H8 x 5 molecules of O2 = 100 molecules of O2
1 molecule of C3H8
3. 6.022x1023 molecules of C3H8 react with _3.011x1024_ molecules of O2 to produce
_1.807x1024_ molecules of CO2 and _2.409 x1024_ molecules of H2O.
Stoichiometric ratios work here too. Ex:
6.022x1023 molecules C3H8 x 5 molecules O2 = 30.11x1023 molecules of O2
1 molecule C3H8 (change to proper scientific notation)
4. 1 mole of C3H8 reacts with __five__ moles of O2 to produce __three__ moles of CO2 and
_four_ moles of H2O. Again, read the coefficients in front of each chemical. These
stoichiometric proportions may represent moles or molecules.
5. 5 moles of C3H8 reacts with _25_ moles of O2 to produce _15_ moles of CO2 and _20_moles
of H2O. Multiply each coefficient above by the stoichiometric ratio. Example shown below.
Ex: 5 moles of C3H8 x 5 moles of O2
1 mole of C3H8
6. 44 g of C3H8 reacts with _160_ grams of O2 to produce _130_ grams of CO2 and _72_ grams
of H2O. Convert grams to moles by dividing by molar mass of propane. Use stoichiometric
proportions as in problem 6. Then multiply the number of moles by the molar mass of each
compound. Example shown below.
44g C3H8 x 1 moles C3H8 x 5 mole O2 x 32 g of O2 = 160g O2
44 g of C3H8 1 mole C3H8 1 mole of O2
**Don’t forget that significant figures matter! These should all have 2 sig figs!**
7. 176 g of C3H8 reacts with __638_ grams of O2 to produce _528_ grams of CO2 and _288_
grams of H2O. Same methods as above in number 6. Three sig figs in this case though.
176 g C3H8 x 1 moles C3H8 x 5 mole O2 x 32.00 g of O2 = 638.40g O2
44.11 g of C3H8 1 mole C3H8 1 mole of O2
2
8. How many moles of carbon dioxide can be produced from the reaction of 12 moles of
propane? Same methods as above in number 6. Two sig figs.
36 moles of carbon dioxide
12 moles of C3H8 x 3 moles of CO2
1 mole of C3H8
9. What mass of oxygen is needed to produce 65 grams of water? 140 g of oxygen
65 g H3O x 1 moles H3O x 5 mole O2 x 32 g of O2 = 144g of O2 (2 sig figs)
18 g H3O 4 mole H3O 1 mole of O2
10. How many moles of carbon dioxide can be produced from the reaction of 225 g of propane?
673 g of carbon dioxide
225 g C3H8 x 1 moles C3H8 x 3 mole CO2 x 44.0 g of CO2 = 673.47g (3 sig figs)
44.1 g of C3H8 1 mole C3H8 1 mole of CO2
11. What mass of H2O is produced from the reaction of 6.3 g of propane? 10.3 g of water
6.3 g C3H8 x 1 moles C3H8 x 4 mole H2O x 18 g of H2O (2 sig figs)
44 g of C3H8 1 mole C3H8 1 mole of H2O
12. How many molecules of H2O are produced when 2 moles of O2 are reacted with excess
propane? Oxygen is the limiting reagent
1 x 1024 molecules of water
2 mol O3 x 4 mole H2O x 6.022x1023 molec. H2O = 9.635 x 1023 molec. (1 sig fig)
5 mole O3 1 mole H2O