Answer :
Answer:
[tex]B_2O_3[/tex]
Explanation:
First, we have to find the reaction:
[tex]B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO[/tex]
The next step is to balance the reaction:
[tex]B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO[/tex]
Now, we have to calculate the molar mass for each compound, so:
[tex]B_2O_3=~69.62~g/mol[/tex]
[tex]C=~12~g/mol[/tex]
[tex]Cl_2=~70.96~g/mol[/tex]
With these values, we can calculate the moles of each compound:
[tex]95.7~g~B_2O_3\frac{1~mol~B_2O_3}{69.62~g~B_2O_3}=1.37~mol~B_2O_3[/tex]
[tex]75.7~g~C\frac{1~mol~C}{112~g~C}=6.30~mol~C[/tex]
[tex]369~g~Cl_2\frac{1~mol~Cl_2}{70.96~g~C}=5.20~mol~Cl_2[/tex]
Now we can divide by the coefficient of each compound in the balanced equation:
[tex]\frac{1.37~mol~B_2O_3}{1}=~1.37[/tex]
[tex]\frac{6.30~mol~C}{3}=~2.1[/tex]
[tex]\frac{5.20~mol~Cl_2}{3}=~1.73[/tex]
The smallest values are for [tex]B_2O_3[/tex], so this is our limiting reagent.
I hope it heps!