Question

If 32.5 g of nitric acid is reacted with 56.3 g of potassium hydroxide, how many grams of potassium nitrate can be formed?

If 25.0 mL of a 7.66 M solution of sulfuric acid is diluted to 300 mL, what is the final concentration of the solution?

The fermentation of a sugar like glucose, C6H12O6, to ethanol, C2H5OH, is readily accomplished by bacteria. The equation is: C6H12O6 ----> 2 C2H5OH + 2 CO2 If we start with 1.25 x 103 g of glucose, how many grams of ethanol can we make?

Answer 1

1)

HNO3   +    KOH -------------------------> KNO3    + H2O

63.0g          56.1g                            101.1g    18 g

32.5g          56.3 g                              ?

63.0 g HNO3 ---------------------------------> 56.1g KOH

32.5 g HNO3 --------------------------------> 32.5 x 56.1 / 63 = 28.94g

actually we need 28.94 g of KOH is needed . but we have 56.3 g KOH . so it is excess reagent . and limiting reagent is HNO3

63.0 g HNO3 ----------------------------> 101.1 g KNO3

32.5 g HNO3 ------------------------> 101.1 x 32.5 / 63 = 52.2 g

mass of potassium nitrate = 52.2 g

2)

C1 V1 = C2 V2 --------------------> dilution formula

7.66 x 25 = C2 x 300

C2 = 0.638 M

final concentration = 0.638 M

3)

from the reaction :

180 g glucose --------------------------------> 92 g ethanol

1.25 x 10^3 g glucose ----------------------> 1.25 x 10^3 x 92 / 180 = 639 g

mass of ethanol made = 639 g