If you combust 7.500g of methanol with an excess of oxygen, a) how many grams and molecules of water will be produced, and b) how many grams and molecules of carbon dioxide will be produced

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

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