Question

How many grams of oxygen does it take to burn 11.4 g of glucose?

What is the theoretical yield in grams of carbon dioxide if 10.9 g of glucose is burned?

Answer 1

1)

Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 11.4 g

mol of C6H12O6 = (mass)/(molar mass)

= 11.4/180.156

= 0.0633 mol

we have the Balanced chemical equation as:

C6H12O6(s) + 6O2(g) -->6CO2(g) + 6H2O(g)

From balanced chemical reaction, we see that

when 1 mol of C6H12O6 reacts, 6 mol of O2 is reacting

mol of O2 reacts = (6/1)* moles of C6H12O6

= (6/1)*0.0633

= 0.3797 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 0.3797*32

= 12.1 g

Answer: 12.1 g

2)

Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 10.9 g

mol of C6H12O6 = (mass)/(molar mass)

= 10.9/180.156

= 0.0605 mol

From balanced chemical reaction, we see that

when 1 mol of C6H12O6 reacts, 6 mol of CO2 is formed

mol of CO2 formed = (6/1)* moles of C6H12O6

= (6/1)*0.0605

= 0.363 mol

Molar mass of CO2 = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 0.363*44.01

= 16.0 g

Answer: 16.0 g