When 50.5 g of silicon dioxide is heated with an excess of carbon, 32.1 g of silicon carbide is produced.?

Question

SiO2(s) + 3 C(s) SiC(s) + 2 CO(g)

What is the percent yield of this reaction?

How many grams of CO gas are made?

Answer 1

First figure out how much SiC SHOULD have been produced by 50.5 g of SiO2. This is a simple mass-mass problem

a. Figure out the molecular mass of SiO2 (add up the atomic masses). Put this number above the SiO2 in the equation.

b. Figure out the molecular mass of SiC (add up the atomic masses). Put this number above the SiC.

c. Look at the coefficients in front of the SiO2 and the SiC: the raio is 1 :1 so the masses you just figured form the gram ratio between SiO2 and SiC. Use this ratio to figure out how much SiC should be produced (theoretical yield)

50.5 g SiO2 x grams SiC/grams SiO2 = grams SiC

Now, take the amount actually produced (32.1g) and divide it by your theoretical yield and change it to a %. This is the percent yield.

Your second question becomes "How many grams of CO gas are produced when 32.1 g of SiC are produced. This is a simple mass-mass problem. Do it the same way you did the first step.

a. You already have the molecular mass of SiC.

b. Get the molecular mass of CO and put it above the CO in the equation.

c. Look at the coefficients: there is a "2" in front of the CO so multiply its molecular mass by 2.

d. Use these two masses to change 32.1 g SiC into grams of CO.

32.1g SiC x grams CO/grams SiC = grams CO.

Answer 2

Looks like a homework problem from a periodic chart determine atomic wts of Si O C Calculate gfws for SiOsub2 CO and SiC. from weights determine gm-moles of Si in each SiOsub2 and SiC. If the gm-moles of SiC is higher than SiOsub2, your teacher pulled a fast one on you or your math is bad. Determine yield by dividing the gm-moles of SiC by gm-moles of SiOsub2. you have yield. If number is calculated is greater than 100 or 1 depending on how you handle your decimals, report 100% From the difference in gm-moles multiple by the gfw of CO to get grams of Carbon Monoxide.