Explanation:
The equation for this reaction is given by;
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
From the reaction;
2 mol of methanol produces 2 mol of CO2 and 4 mol of water
In terms of mass;
Mass = Molar mass * Number of mol
Methanol;
Mass = 32 g/mol * 2 mol = 64 g
CO2;
Mass = 44 g/mol * 2 mol = 88g
H2O;
Mass = 18 g/mol * 4 mol = 72 g
This means;
64g of methanol produces 88g of CO2 and 72g of water
a) how many grams and molecules of water will be produced
64g of Methanol = 72g of Water
7.5g would produce xg of Water
X = (7.5 * 72) / 64 = 8.4375 g of water
Number of moles = Mass / Molar mass = 8.4375 / 18 = 0.46875 mol
Molecules = moles * Avogadro number = 0.46875 * 6.022 * 10^23 = 2.82 * 10^23 molecules
b) how many grams and molecules of carbon dioxide will be produced
64g of Methanol = 88g of CO2
7.5g would produce xg of CO2
X = (7.5 * 88) / 64 = 10.3125 g of CO2
Number of moles = Mass / Molar mass = 10.3125 / 44 = 0.2344 mol
Molecules = moles * Avogadro number = 0.2344 * 6.022 * 10^23 = 1.411 * 10^23 molecules
