Question

2) Balance C4H10 + O2 → CO2 + H2O & solve:

a) If 7.54 mol of CO2 are produced, how many moles of O2 reacted?

b)If 9.75 kg of H2O are produced, how many moles of C4H10 reacted?

c) If 3.4500 moles of O2 react, how many grams of CO2 are produced?

d)If 520.30 grams of C4H10 react, how many mg of H2O are produced?

e) If 4.28 mol H2O are produced, how many molecules of O2 reacted?

Answer 1

2)

Balanced equation is

C4H10 + 6.5 O2 $\to$ 4 CO2 + 5 H2O

a) mole ratio of CO2 and O2 = 4 : 6.5

then ,

when 7.54 mol CO2 is formed , then moles of O2 reacted

= 7.54 CO2 $\times$

6.5 mole 02 4 mole CO2

= 12.25 mole

b)

9.75 Kg of H2O = 9.75 $\times$ 1000 = 9750 g

moles of water

= ( 9750/18)

= 541.66

now, mole ratio of H2O and C4H10

= 5 : 1

then , moles of C4H10 reacted = (541.66/5) = 108.33

c)

Mole ratio of CO2 and O2 = 4 : 6.5

moles of CO2 formed when 3.45 moles O2 reacted

= 3.45 $\times$

4 mole CO2 6.5 mole 02

= 2.12

then , grams of CO2 formed = moles $\times$ molar mass = 2.12 (mol) $\times$ 44 (g/mol) = 93.28 g

d)

Molar mass of C4H10 = 58 g/mol

From balanced reaction

58 g C4H10 reacts to give 5 $\times$ 18 = 90 g water

then, 520.30 grams of C4H10 reacts to give

520.30  $\times$ (  $\frac{90}{58}$ ) g water

= 807.362 g water

= (807.362 $\times$ 1000) mg water

= 807362 mg water .

e)

mole ratio of water and O2

= 5 : 6.5

then , when 4.28 g water is formed , moles of O2 reacted

= 4.28 mole H2O  $\times$$\frac{6.5\: mole\: O_{2}}{5 \: mole\: H_{2}O}$

= 5.56 mole O2

now, 1 mole oxygen contains  6.02$\times$ 10 23   number of molecules

then, molecules of O2 reacted

=6.02$\times$ 10 23 $\times$ 5.56

= 3.35 $\times$ 1024