How many grams of glucose (C6H12O6, 180.16 g/mol) would be required to prepare 2.50 L of a 0.278M glucose solution

125.21 g of glucose

Explanation:

From the question given above, the following data were obtained:

Molar mass of C6H12O6 = 180.16 g/mol

Volume of solution = 2.50 L

Molarity of C6H12O6 = 0.278 M

Mass of C6H12O6 =.?

Next, we shall determine the number of mole of C6H12O6 in the solution. This can be obtained as follow:

Volume of solution = 2.50 L

Molarity of C6H12O6 = 0.278 M

Number of mole of C6H12O6 =?

Molarity = mole /Volume

0.278 = mole of C6H12O6 / 2.50

Cross multiply

Mole of C6H12O6 = 0.278 × 2.50

Mole of C6H12O6 = 0.695 mole

Finally, we shall determine the mass of glucose, C6H12O6, required to prepare the solution as follow:

Molar mass of C6H12O6 = 180.16 g/mol

Mole of C6H12O6 = 0.695 mole

Mass of C6H12O6 =.?

Mole = mass /molar mass

0.695 = mass of C6H12O6 / 180.16

Cross multiply

Mass of C6H12O6 = 0.695 × 180.16

Mass of C6H12O6 = 125.21 g

Thus, 125.21 g of glucose, C6H12O6 is needed for the preparation of the solution.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g