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). Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
Calculate the mass of butane needed to produce 71.6 of carbon dioxide

Answer 1

ok so first lets calculate amount of butane which is in 1,77 g:
n=m/M = 1,77 g/58,124 gmol-1=0,03045 mol
now lets calculate mass of water from reaction ratio:
n(H2O)/n(C4H10) = 10/2 = 5 --> n(H2O) = 5 n(C4H10) = 5* 0,03045 mol
n(H2O)= 0,15225 mol
n=m/M --> m=n*M
m(H2O)= n(H2O) * M(H2O)
m(H2O)= 0,15225 mol * 18 gmol-1 = 2,74 g

that's first part, now second part:

lets again calculate amount from mass (at least i guess that's mass):
n=m/M = 71,6 g / 44 gmol-1 = 1,63 mol
now again lets calculate mass of butane from reaction ratio:
n(C4H10)/n(CO2)=2/8 = 1/4 --> n(C4H10) = 1/4 * n(CO2)
n(C4H10) = 1/4 * 1,63 mol = 0,4075 mol
and again lets calculate mass:
m= n*M = 0,4075 mol * 58,124 gmol-1 = 23,68 g

Answer 2

Thank you s much.