mass of carbon dioxide produced when 0.85 g of butane reacts with oxygen 2 C4H10 + 13 O2 ® 8 CO2 + 10 H2O?

Answer 1

i got 176.04 grams of CO2

Answer 2

Find the molar mass of butane, which is the mass of 1 mole of butane. Then, with that, find out how many moles .85 grams of butane is.

Once you know how many moles .85 is, then you look at the ratio in the balanced equation...

In the equation it says 2 moles of C4H10 used per every 8 moles of CO2 produced.

So, however many moles .85 grams of butane is, multiply it by 4, and that's how many moles of CO2 there are...

Then, you find the molar mass of CO2, and multiply it by how many moles of CO2 there are, and that will be the mass of CO2 produce through the reaction of Butane and oxygen.

Answer 3

{a million} 2 moles of butane require 8 moles of carbon(iv)oxide then 5 moles of butane will require (5x8/2)moles of carbon(iv)oxide it is 20moles. {2} 2 moles of butane require 10 moles of water then 5 moles will require (5x10/2)moles =25molesbut mass=molar mass X no. of moles=18 X 25=450g the place 18= molar mass of water