1)If 4.09 moles of C5H12 reacts with excess O2, how many moles of CO2 will be pr
ID: 540099 • Letter: 1
Question
1)If 4.09 moles of C5H12 reacts with excess O2, how many moles of CO2 will be produced in the following combustion reaction?
2)The first step in the reaction of Alka-Seltzer with stomach acid consists of one mole of sodium bicarbonate (NaHCO3) reacting with one mole of hydrochloric acid (HCl) to produce one mole of carbonic acid (H2CO3), and one mole of sodium chloride (NaCl).
Using this chemical stoichiometry, determine the number of moles of carbonic acid that can be produced from 5 mol of NaHCO3 and 8 mol of HCl.
How much excess reactant remains after the reaction?
Explanation / Answer
1) c5h12(g) + 8O2(g) ----> 5CO2(g) + 6H2O(l)
from balanced equation
1 mol C5H12 = 5 mol CO2
No of mol of C5H12 reacted = 4.09 moles
No of mol of CO2 produced = 4.09*5 = 20.45 mol
2) NaHCO3 + HCl ---> H2CO3 + NaCl
from equation : 1 mol NaHCO3 = 1 mol HCl
from 5 mol of NaHCO3 and 8 mol of HCl.
LIMITING REACTANT = NaHCO3
no of mol of H2CO3 produced = 5 mol
No of mol of excess HCl left in the reaction = 8-5 = 3 mol
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