If you add 14g of Hydrogen to excess oxygen, how much water should form?
H2 = 2 grams/mol
O = 16 grams/mol
H2O = 18 grams /mol
so
14 grams H2 is 7 mol
that means we get 7 mol of water
7 * 18 = 126 grams
adddfgggs
To determine how much water should form when adding 14g of Hydrogen to excess oxygen, we can use the equation:
2H2 + O2 → 2H2O
The molar mass of hydrogen (H2) is 2g/mol. Therefore, 14g of hydrogen is equivalent to:
14g / 2g/mol = 7 mol of hydrogen
According to the balanced equation, for every 2 moles of hydrogen, 2 moles of water will form. Thus, with 7 moles of hydrogen, the amount of water formed is:
(7 mol H2) x (2 mol H2O / 2 mol H2) = 7 mol H2O
The molar mass of water (H2O) is 18g/mol. Therefore, the mass of water formed is:
7 mol H2O x 18g/mol = 126g
So, 126g of water should form when adding 14g of hydrogen to excess oxygen.
To determine how much water should form when 14g of hydrogen is added to excess oxygen, we need to consider the balanced chemical equation for the reaction:
2H₂ + O₂ → 2H₂O
From the equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.
To calculate the moles of hydrogen, we use the molar mass of hydrogen (1g/mol). So, 14g of hydrogen is equal to 14 moles.
Using the mole ratio from the balanced equation, we can determine the moles of water produced. Since the ratio is 2:2, for every 2 moles of hydrogen, 2 moles of water are formed.
Therefore, the moles of water produced would also be 14 moles.
To convert moles of water to grams, we use the molar mass of water (H₂O), which is (2 x 1.008) + 16.00 = 18.015 g/mol.
From the moles of water produced, we can calculate the mass:
Mass = Moles x Molar mass
Mass = 14 moles x 18.015 g/mol
Mass = 252.21 g
Therefore, when 14g of hydrogen is added to excess oxygen, approximately 252.21g of water should form.