How many moles of O₂ react with or are produced by 5.67 moles of the first reactant shown in each balanced equation?

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7. How many moles of O₂ react with or are produced by 5.67 moles of the first reactant shown in each
balanced
equation?
4CO2 + 6H₂O
a.
b.
C.
b.
C.
b.
C.
d.
e.
2C2H6 + 702
2KCIO3
8. How many grams of the second reactant is needed to react with 18.3 grams of the first reactant in each
balanced equation?
a.
4Fe + 30₂2
9 c.
9 d.
9 e.
a.
P4 + 6C1₂
b.
P4O10+ 6H2O
2KC1+30₂
203 + 3CO
P4O10+ 6H2O
9 a.
9 b.
→2Fe₂O3
9. Using the balanced equations below, complete 9.a through 9.d:
a.
4PC13
4H3PO4
How many grams of the first product can be produced from 15.0 grams of the first reactant
2Fe + 3CO2
and 15.0 grams of the second reactant?
Which reactant is the limiting reactant?
Which reactant is the excess reactant?
4H3PO4
How many grams of the excess reactant are reacted?
How many grams of the excess reactant are left unreacted?
10. A student carried out the reaction shown by reacting 7.38g P4 with excess Cl₂ and obtained 35.0g of
PC13.
P4 + 6Cl24PCL3
What is the theoretical yield of PC13?
Calculate the % yield of PC13.
Transcribed Image Text:7. How many moles of O₂ react with or are produced by 5.67 moles of the first reactant shown in each balanced equation? 4CO2 + 6H₂O a. b. C. b. C. b. C. d. e. 2C2H6 + 702 2KCIO3 8. How many grams of the second reactant is needed to react with 18.3 grams of the first reactant in each balanced equation? a. 4Fe + 30₂2 9 c. 9 d. 9 e. a. P4 + 6C1₂ b. P4O10+ 6H2O 2KC1+30₂ 203 + 3CO P4O10+ 6H2O 9 a. 9 b. →2Fe₂O3 9. Using the balanced equations below, complete 9.a through 9.d: a. 4PC13 4H3PO4 How many grams of the first product can be produced from 15.0 grams of the first reactant 2Fe + 3CO2 and 15.0 grams of the second reactant? Which reactant is the limiting reactant? Which reactant is the excess reactant? 4H3PO4 How many grams of the excess reactant are reacted? How many grams of the excess reactant are left unreacted? 10. A student carried out the reaction shown by reacting 7.38g P4 with excess Cl₂ and obtained 35.0g of PC13. P4 + 6Cl24PCL3 What is the theoretical yield of PC13? Calculate the % yield of PC13.
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