What mass of boron sulfide must be processed with 2.1 x 10 4g of carbon to yield 3.11 x 10 4 g of boron and 1.47 x 10 5 g of carbon sulfide?

The reaction between boron sulfide and carbon is given as:

2B2S3 + 3C → 4B + 3CS2

As per the law of conservation of mass, for any chemical reaction the total mass of reactants must be equal to the total mass of the products.

Given dа ta:

Mass of C = 2.1 * 10^ 4 g

Mass of B = 3.11*10^4 g

Mass of CS2 = 1.47*10^5

Mass of B2S3 = ?

Now based on the law of conservation of mass:

Mass of B2S3 + mass C = mass of B + mass of CS2

Mass of B2S3 + 2.1 * 10^ 4 = 3.11*10^4 + 1.47*10^5

Mass of B2S3 = 15.7 * 10^4 g