What coefficient would the O2 have after balancing C4H10+ 02 → CO2 + H2O?
A. 5
B. 2.5
C. 4
D. 6.5

option D. 6.5

Explanation:

C4H10 + 02 → CO2 + H2O

To balance the the above equation, add 2 in front of C4H10, 13 in front of O2, 8 in front of CO2 and 10 in front of H2O i.e

2C4H10 + 13O2 → 8CO2 + 10H2O

Next divide through by 2,we have

C4H10 + 6.5O2 → 4CO2 + 5H2O

From the above illustration, we see that the coefficient of O2 is 6.5

true

explanation:

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g