Silicon carbide (SiC) is an important ceramic material made by reacting sand (silicon dioxide, SiO2) with powdered carbon at a high temperature. Carbon monoxide is also formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is recovered. What is the percent yield of SiC from this process

Percent yield of SiC is 77.0%.

Explanation:

Balanced reaction:

Molar mass of SiC = 40.11 g/mol

Molar mass of = 60.08 g/mol

So, 100.0 kg of = moles of = 1664 moles of

According to balanced equation, 1 mol of produces 1 mol of SiC

Therefore, 1664 moles of produce 1664 moles of SiC

Mass of 1664 moles of SiC = = 66743g = 66.74 kg (4 sig. fig.)

Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]%

                                 = %

                                 = 77.0%

it's filtration, but this is an assumption so my apologies if i got this

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole