How many grams of water, H2O, can be formed from the reaction of 8.9 grams of 2-propanol, C3H8O?
To determine the grams of water that can be formed from the reaction of 8.9 grams of 2-propanol (C3H8O), we need to consider the balanced chemical equation and the molar ratios.
The balanced chemical equation for this reaction is:
C3H8O + 4O2 → 3CO2 + 4H2O
From the equation, we can see that for every 1 mole of C3H8O, 4 moles of H2O are formed.
To calculate the moles of 2-propanol, we divide the given mass (8.9 grams) by the molar mass of 2-propanol (C3H8O).
The molar mass of C3H8O:
3(C) + 8(H) + 1(O) = 44.10 g/mol
Moles of C3H8O = 8.9 g / 44.10 g/mol = 0.2018 mol (rounded to four decimal places)
Since the molar ratio of C3H8O to H2O is 1:4, we can calculate the moles of water formed by multiplying the moles of C3H8O by the molar ratio of water.
Moles of H2O = 0.2018 mol (C3H8O) x 4 mol (H2O) / 1 mol (C3H8O) = 0.8072 mol (rounded to four decimal places)
To convert moles of water to grams, we multiply the moles of water by the molar mass of water (H2O).
Molar mass of H2O:
2(H) + 1(O) = 18.02 g/mol
Grams of H2O = 0.8072 mol (H2O) x 18.02 g/mol = 14.55 grams (rounded to two decimal places)
Therefore, from the reaction of 8.9 grams of 2-propanol (C3H8O), approximately 14.55 grams of water (H2O) can be formed.