Question

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

a) Calculate the mass of water produced when 2.04 g of butane reacts with excess oxygen.

b) Calculate the mass of butane needed to produce 72.8 g of carbon dioxide.

Answer 1

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

as per the above eq. 2moles of butane gives ..................10 moles water

       no. of moles of butane given = 2.04/58     = 0.035 moles

now      

2moles of butane gives ..................10 moles water

0.035 moles of C4H10 .....................?

= 0.035*10/2 = 0.175 moles of water

weight of water = no.of moles *mol.wt = 0.175*18      = 3.15 gm.

.................................................................................................................................

b)

as per the above eq 8 moles of CO2 formed from ............ 2moles of butane

   given 72.8 gm. of CO2

no.of moles of CO2 = 72.8/44 = 1.654 moles

now

8 moles of CO2 formed from ............ 2moles of butane

1.654 moles of CO2 ................................?

   = 1.654*2/8   = 0.4135 moles of butane

wt of butane = 0.4135*58 = 23.98 gm.

23.98 gm. of Butane reuired.