Question

Answer 1

\(n_{C_2H_2}=\dfrac{130}{26}=5\left(mol\right)\)

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

  5            12,5                             ( mol )

\(m_{O_2}=12,5.32=400\left(g\right)\)

Answer 2

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{130}{26}=5\left(mol\right)\\ n_{O_2}=\dfrac{5}{2}.5=12,5\left(mol\right)\\ \Rightarrow m_{O_2}=12,5.32=400\left(g\right)\)