Đổi 12,47(kg) = 12470 (g)
\(n_{C4H10}=\frac{12470}{58}=215\left(mol\right)\)
\(2C_4H_{10}+9O_2\underrightarrow{^{to}}8CO_2+10H_2O\)
a, Theo PT:
\(n_{O2}=\frac{9}{2}n_{C4H10}=\frac{9}{2}.215=967,5\left(mol\right)\)
\(\Rightarrow V_{kk}=5.22,4.967,5=108360\left(l\right)\)