Answer:
A. Amount of SiO₂ = 102941.25 moles of SiO₂
Amount of C = 308823.75 moles of C
B. 0.712 metric tons of coal is required for every metric ton of sand
Note: The question is incomplete. the complete question is given below;
Silicon carbide is an abrasive used in the manufacture of grinding wheels. A company is investigating whether they can be more efficient in the construction of such wheels by making their own SiC via the reaction.
SiO₂ + 3C = SiC + 2CO
a. If this reaction consistently has a yield of 83%, what is the minimum amount of both silicon dioxide and carbon needed to produce 3500 kg of SiC for the manufacture of cutting wheels?
Amount of silicon dioxide = ? mol
Amount of carbon = ? mol
b. The silicon dioxide is to be obtained from sand and the carbon is derived from coal. If the available sand is 93% SiO2 by weight and the coal is 75% C by weight, what mass of coal is needed for each metric ton of sand used?
Mass of coal = ? metric tons
Step-by-step explanation:
A. First the theoretical yield from the equation is determined
100/85* 3500 kg = 4117.65 kg = 4117650 g
molar mass of silicon dioxide = 28 + 2*16 = 60 g/mol; molar mass of carbon = 12g/mol; molar mass f SiC = 28 + 12 = 40 g/mol
40 g SiC is produced by 60 g of SiO₂
4117650 g of SiC will be produced by 4117650 * 60/40 g of SiO₂ = 6176475 g of SiO₂ = 6176.475 kg of SiO₂
Amount in moles = 6176475 g / 60 g/mol = 102941.25 moles of SiO₂
40 g SiC is produced by 3 * 12 g of C = 36 g of C
4117650 g of SiC will be produced by 4117650 * 36/40 g of C = 3705885 g of C = 3705.885 kg of C
Amount in moles = 3705885 g / 12 g/mol = 308823.75 moles of C
B. I metric tonne = 1000 kg
amount of sand required = 100/93 * 6176.475 kg = 6641.37 kg = 6.64 tonnes
amount of coal required = 100/75 * 3705.885 kg = 4941.18 kg = 4.94 tonnes
ratio of coal to sand = 4.94/6.64 = 0.712
Therefore, 0.712 metric tons of coal is required for every metric ton of sand