a,
$2C_4H_{10}+ 13O_2\buildrel{{t^o}}\over\to 8CO_2+10H_2O$
b,
$D_{H_2O}=1\Rightarrow m_{H_2O}=1000g$
Nâng $1000g$ nước lên $1^oC$ cần $1000.4,184=4184(J)=4,184(kJ)$
$\Delta t=100-25=75^oC$
$\Rightarrow Q=4,184.75=313,8(kJ)$
Do hao phí $10%Q$ nên cần tạo nhiệt lượng $313,8.110\%=345,18(kJ)$
$\Rightarrow n_{C_4H_{10}}=\dfrac{2870,2}{345,18}=8,3(mol)$
$\Rightarrow V_{C_4H_{10}}=8,3.22,4=185,92l=1,8592(m^3)$