Question

How many grams of sugar will be needed to make10 g С2Н5ОН
C6H1206 → 2C2H5OH + 2C02

Answer 1

To calculate the amount of sugar (C6H12O6) needed to produce 10g of C2H5OH (ethyl alcohol), we must first calculate the molar mass of C2H5OH and then use stoichiometry to find the amount of sugar required.

1. Calculate the molar mass of C2H5OH:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol

Molar mass of C2H5OH = (2*12.01) + (6*1.01) + 16.00 = 46.07 g/mol

2. Use the stoichiometry of the reaction to find the amount of sugar needed:
1 mol of C6H12O6 produces 2 mol of C2H5OH
From the balanced equation: 1 mol of C6H12O6 = 2 mol of C2H5OH

Now, calculate how many moles of C2H5OH are produced from 10g:
10g / 46.07 g/mol = 0.22 moles of C2H5OH

Since the ratio of C6H12O6 to C2H5OH is 1:2, we can say that for every 1 mole of C6H12O6, 2 moles of C2H5OH are produced. Therefore, we would need double the amount of sugar than the moles of C2H5OH.

0.22 moles of C2H5OH * 2 = 0.44 moles of C6H12O6

3. Calculate the mass of sugar needed:
Mass of sugar = moles of C6H12O6 * molar mass of C6H12O6
Mass of sugar = 0.44 moles * 180.16 g/mol = 79.27 grams

Therefore, 79.27 grams of sugar would be needed to produce 10g of C2H5OH from C6H12O6.