Answer: 37.44 g
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles and weighs equal to the molecular mass.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
[tex]\text{Number of moles of methanol}=\frac{25g}{32g/mol}=0.78moles[/tex]
[tex]2CH_3OH+3O_2\rightarrow 2CO_2+4H_2O[/tex]
According to stoichiometry:
2 moles of methanol react with 3 moles of oxygen
Thus 0.78 moles of methanol will react with=[tex]\frac{3}{2}\times 0.78=1.17[/tex] moles of oxygen
Mass of oxygen =[tex]moles\times {\text {molar mass}}=1.17\times 32=37.44g[/tex]
Thus 37.44 g of oxygen will react with 25.0 g of [tex]CH_3OH[/tex]
