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 via the following reaction. (a) If this reaction consistently has a yield of , what is the minimum amount (in ) of both silicon dioxide and carbon needed to produce of for the manufacture of cutting wheels? silicon dioxide kg carbon kg (b) The silicon dioxide is to be obtained from sand and the carbon is derived from coal. If the available sand is by weight and the coal is by weight, and assuming the other material does not interfere with the intended reaction, what mass of coal (in ) is needed for each metric ton of sand used? () kg

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

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

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.