Liquid benzene ( C 6 H 6) bums in oxygen according to 2 C 6 H 6( l )+15 O 2( g ) longrightarrow 12 CO 2( g )+6 H 2 O ( g ) How many litres of O 2 at STP are needed to complete the combustion of 39 g of liquid benzene?

Question

Liquid benzene ( C 6 H 6) bums in oxygen according to 2 C 6 H 6( l )+15 O 2( g ) longrightarrow 12 CO 2( g )+6 H 2 O ( g ) How many litres of O 2 at STP are needed to complete the combustion of 39 g of liquid benzene? (A) 74 L (B) 11.2 L (C) 22.4 L (D) 84 L

Tardigrade Admin · Accepted Answer

2 C 6 H 6( l )+15 O 2( g ) longrightarrow 12 CO 2( g )+6 H 2 O ( g ) 2 × 78 g 15 × 22.4 L From equation 15 × 22.4 L of O 2 is required for =156 g of benzene i.e., 156 g benzene for complete combustion required O 2( STP )=15 × 22.4 L 39 g benzene for complete combustion required O 2( STP )=(15 × 22.4/156) × 39=84 L of O 2

Q. Liquid benzene (C6​H6​) bums in oxygen according to 2C6​H6​(l)+15O2​(g)⟶12CO2​(g)+6H2​O(g) How many litres of O2​ at STP are needed to complete the combustion of 39g of liquid benzene?

74 L

11.2 L

22.4 L

84 L

2C6​H6​(l)+15O2​(g)⟶12CO2​(g)+6H2​O(g) 2×78g15×22.4L From equation 15×22.4L of O2​ is required for =156g of benzene i.e., 156g benzene for complete combustion required O2​(STP)=15×22.4L 39g benzene for complete combustion required O2​(STP)=15615×22.4​×39=84L of O2​

Q. Liquid benzene $(C_{6} H_{6})$ bums in oxygen according to $2 C_{6} H_{6} (l) + 15 O_{2} (g) ⟶ 12 C O_{2} (g) + 6 H_{2} O (g)$ How many litres of $O_{2}$ at STP are needed to complete the combustion of $39 g$ of liquid benzene?