Question

Determine the limiting reactant in a mixture containing 95.7 g of B2O3, 75.7 g of C, and 369 g of Cl2. Calculate the maximum mass (in grams) of boron trichloride, BCl3, that can be produced in the reaction. The limiting reactant is:

Answer 1

`B_2O_3`

Step-by-step explanation:

First, we have to find the reaction:

`B_2O_3~+~C~+~Cl_2~->~BCl_3~+~CO`

The next step is to balance the reaction:

`B_2O_3~+~3C~+~3Cl_2~->~2BCl_3~+~3CO`

Now, we have to calculate the molar mass for each compound, so:

`B_2O_3=~69.62~g/mol`

`C=~12~g/mol`

`Cl_2=~70.96~g/mol`

With these values, we can calculate the moles of each compound:

`95.7~g~B_2O_3(1~mol~B_2O_3)/(69.62~g~B_2O_3)=1.37~mol~B_2O_3`

`75.7~g~C(1~mol~C)/(112~g~C)=6.30~mol~C`

`369~g~Cl_2(1~mol~Cl_2)/(70.96~g~C)=5.20~mol~Cl_2`

Now we can divide by the coefficient of each compound in the balanced equation:

`(1.37~mol~B_2O_3)/(1)=~1.37`

`(6.30~mol~C)/(3)=~2.1`

`(5.20~mol~Cl_2)/(3)=~1.73`

The smallest values are for
`B_2O_3`, so this is our limiting reagent.

I hope it heps!