In photosynthesis, green plants convert CO2 and H2O to glucose, CHs How many gra
ID: 1031404 • Letter: I
Question
In photosynthesis, green plants convert CO2 and H2O to glucose, CHs How many grams of Co are required to produce 5.1 g of glucose? 4-61 6112 6 esis 12 Glucose Iron ore is converted to iron by heating it with coal (carbon), and oxygen according to the following equation: X 4-62 2Fe203(s) + 6C(s) +302(g)4Fe(s) + 6CO2(g) If the process is run until 3940. g of Fe is produced how many grams of CO2 will also be produced? Given the reaction in Problem 4-62, how many grams of C are necessary to react completely with 0.58 g of Fe2O3? 4-63Explanation / Answer
4 - 61
molar mass of glucose = 180.156 gm/mol
no. of mole = mass of compound in gm / molar mass
no. of mole of glucose = 5.1 / 180.156 = 0.0283 mole
According to reaction 6 mole CO2 produce 1 mole glucose molar retio between CO2 to glucose is 6:1 therefore to prepare 0.0283 mole glucose required CO2 = 0.0283 X 6 = 0.1698 mole
0.1698 mole CO2 required
molar mass of CO2 = 44.01 g/mol
gm of compound = no. of mole X molar mass
gm of CO2 required = 0.1698 X 44.01 = 7.47 gm
gm of CO2 required = 7.47 gm
4 - 62
molar mass of Fe =55.845g/mol
no. of mole = mass of compound in gm / molar mass
no. of mole of glucose = 3940/55.845= 70.55 mole
According to reaction 6 mole CO2 produce when 4 mole of Fe produced molar retio between CO2 to Fe is 3:2 therefore 70.55 mole Fe produce then 70.55 X 3 / 2 = 105.825 mole of CO2 produced
105.825mole of CO2 produced
molar mass of CO2 = 44.01 g/mol
gm of compound = no. of mole X molar mass
gm of CO2 required = 105.825 X 44.01= 4657.36 gm
gm of CO2 produced =4657.36 gm
4 - 63
molar mass of Fe2O3 = 159.69 g/mol
no.of mole = gm of compound / molar mass
no. of mole of Fe2O3 = 0.58 / 159.69 = 0.003632 mole
According to reaction to react with 2 mole Fe2O3 required C is 6 molar thus molar retio between Fe2O3 to C is 1:3 therefore to react with 0.003632 mole of Fe2O3 required C= 0.03632 X 3 = 0.010896 mole of C
molar mass of C = 12.0107 g/mol
gm of compound = no. of mole X molar mass
gm of C required = 0.010896 X 12.0107 = 0.13 gm
0.13 gm C required