Question

I need a little help with this problem. Calculate how many moles of BCl3 are produced from the reaction of 61.0g of B203?

Answer 1

\[B _{2}O _{3} (s) + 3C (s) + 3Cl _{2} (g) \rightarrow 2BCl _{3} (g) +3CO (g)\]

Answer 2

I know that my conversion is wrong somehow; I just need someone to confirm where.
\[\frac{ 61g B _{2}O _{3} }{ 1 } X \frac{ 1 mol B _{2}O _{3} }{ 69.62g B _{2}O _{3} } X \frac{ 2 mol BCl _{3} }{ 1 mol B _{2} O _{3}}=1.75 mol BCl _{3}\]

Answer 3

This is the exact question.