Calculate the mass of water produced when 3.94 g of butane reacts with excess oxygen.

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation:
2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

mols C4H10 = 3.94/molar mass = ?
For every 2 mols C4H10 you get 10 mol H2O; therefore, mols H2O = ?
Convert mols H2O to grams. g = mols x molar mass = ?
Post your work if you get stuck.

To calculate the mass of water produced, we first need to find the mole ratio between butane and water using the balanced chemical equation.

From the balanced equation, we can see that 2 moles of butane produce 10 moles of water.

1) Convert the mass of butane to moles:
molar mass of butane (C4H10) = (4 * atomic mass of Carbon) + (10 * atomic mass of Hydrogen)
molar mass of butane (C4H10) = (4 * 12.01 g/mol) + (10 * 1.01 g/mol)
molar mass of butane (C4H10) = 58.12 g/mol

moles of butane = mass of butane / molar mass of butane
moles of butane = 3.94 g / 58.12 g/mol
moles of butane = 0.068 mol

2) Calculate the moles of water produced using the mole ratio:
moles of water = moles of butane * (moles of water / moles of butane)
moles of water = 0.068 mol * (10 mol water / 2 mol butane)
moles of water = 0.34 mol

3) Convert moles of water to mass:
mass of water = moles of water * molar mass of water
molar mass of water (H2O) = (2 * atomic mass of Hydrogen) + (1 * atomic mass of Oxygen)
molar mass of water (H2O) = (2 * 1.01 g/mol) + (1 * 16.00 g/mol)
molar mass of water (H2O) = 18.02 g/mol

mass of water = 0.34 mol * 18.02 g/mol
mass of water = 6.13 g

Therefore, when 3.94 g of butane reacts with excess oxygen, the mass of water produced is 6.13 g.

To calculate the mass of water produced when 3.94 g of butane reacts with excess oxygen, we need to use the stoichiometry of the balanced chemical equation.

First, we need to determine the molar mass of butane (C4H10), which is 58.12 g/mol (4 * atomic mass of carbon + 10 * atomic mass of hydrogen).

Next, we need to calculate the moles of butane used. We can use the formula:
moles = mass / molar mass.

moles of butane = 3.94 g / 58.12 g/mol = 0.0678 mol of butane.

According to the balanced chemical equation, the stoichiometric ratio of butane to water is 2:10. This means that for every 2 moles of butane, we get 10 moles of water.

moles of water = (moles of butane) * (moles of water / moles of butane ratio) = 0.0678 mol * (10 mol/ 2 mol) = 0.339 mol of water.

Finally, we can calculate the mass of water produced using the formula:
mass = moles * molar mass.

mass of water = 0.339 mol * 18.015 g/mol = 6.10 g.

Therefore, when 3.94 g of butane reacts with excess oxygen, the mass of water produced is approximately 6.10 g.