What weight of CO is required to form Re2(CO)_{10} from 2.42 g of Re2O7 according to the given unbalanced reaction?

Question

What weight of $CO$ is required to form $Re _{2}( CO )_{10}$ from $2.42 g$ of $\operatorname{Re}_{2} O _{7}$ according to the given unbalanced reaction?

$\operatorname{Re}_{2} O _{7}+ CO \longrightarrow Re _{2}( CO )_{10}+ CO _{2} \quad$ (Atomic mass of $Re =186 u$ )

Answer 1

2.38 g weight of $CO$ is required to form $Re _{2}( CO )_{10}$ from $2.42 g$ of $\operatorname{Re}_{2} O _{7}$ according to the given unbalanced reaction.