2C4H10(g) + 1302 ===> 8CO2(g) + 10H2O(g) What mass of carbon dioxide (CO2) would be produced when 10.0 g of butane (C4H10) reacts with
excess oxygen?

30.34g (corrected to 4 significant figures).

Explanation:

Take the atomic mass of C=12.0, H=1.0, O=16.0.

no. of moles = mass / molar mass

So, no. of moles of butane reacted = 10 / (12x4 + 1x10)

= 0.172414 mol

Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.

From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,

meaning 1 mole of butane gives 4 moles of CO2.

Using this ratio,

we can deduce that the no. of moles of CO2 produced =  0.172414 x 4

=0.689655 mol

As mass = no. of moles x molar mass

mass of CO2 produced =   0.689655 x (12.0+16.0x2)

=30.34g (corrected to 4 significant figures).