how many grams of water can be formed when 28.7 grams of CH4 reacts with 45.8 grams of O2

See your problem above. This one is a limiting reagent problem because BOTH reactants are given. What you do is to solve the stoichiometry problem twice, once with the CH4 amount and once with the O2 amount. You will get two answers for the moles H2O formed and the correct answer is the the smaller of the two.

To determine the amount of water formed when CH4 reacts with O2, we need to first write the balanced chemical equation for the reaction:

CH4 + 2O2 → CO2 + 2H2O

From the balanced equation, we can see that one mole of CH4 reacts with two moles of O2 to produce two moles of water (H2O).

Step 1: Calculate the moles of CH4:

Molar mass of CH4 (carbon + 4 hydrogen) = 12.01 g/mol + (4 * 1.01 g/mol) = 16.05 g/mol

Moles of CH4 = Mass of CH4 / Molar mass of CH4
= 28.7 g / 16.05 g/mol
≈ 1.79 mol

Step 2: Calculate the moles of O2:

Molar mass of O2 (oxygen) = 2 * 16.00 g/mol = 32.00 g/mol

Moles of O2 = Mass of O2 / Molar mass of O2
= 45.8 g / 32.00 g/mol
≈ 1.43 mol

Step 3: Determine the limiting reactant:

The limiting reactant is the one that is completely consumed in the reaction, limiting the amount of product that can be formed. We determine the limiting reactant by comparing the number of moles of each reactant with the stoichiometric ratio in the balanced equation.

From the balanced equation: 1 mole of CH4 reacts with 2 moles of O2

Since we have 1.79 moles of CH4 and 1.43 moles of O2, we can see that O2 is the limiting reactant because we have fewer moles of O2 compared to the stoichiometric ratio.

Step 4: Calculate the moles of water formed:

From the balanced equation, we know that 2 moles of water (H2O) are produced from 1 mole of CH4.

Moles of water formed = 2 * Moles of limiting reactant (O2)
= 2 * 1.43 mol
≈ 2.86 mol

Step 5: Calculate the mass of water formed:

Molar mass of water (H2O) = 2 * 1.01 g/mol + 16.00 g/mol = 18.02 g/mol

Mass of water formed = Moles of water formed * Molar mass of water
= 2.86 mol * 18.02 g/mol
≈ 51.52 g

Therefore, approximately 51.52 grams of water can be formed when 28.7 grams of CH4 reacts with 45.8 grams of O2.

To find out the number of grams of water formed when CH4 reacts with O2, we first need to determine the balanced chemical equation for the reaction.

The balanced chemical equation for the combustion of CH4 (methane) can be written as:

CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that for every mole of CH4, we produce 2 moles of H2O. Now, we need to calculate the number of moles of CH4 and O2 present in the reaction.

To calculate the number of moles, we use the formula:

moles = mass / molar mass

The molar mass of CH4 (methane) = 12.01 g/mol (carbon) + 4(1.01 g/mol) (hydrogen) = 16.05 g/mol

The molar mass of O2 (oxygen) = 2(16.00 g/mol) = 32.00 g/mol

Using the formula, we can find the number of moles of each reactant:

moles of CH4 = 28.7 g / 16.05 g/mol ≈ 1.79 mol
moles of O2 = 45.8 g / 32.00 g/mol ≈ 1.43 mol

Now that we know the moles of CH4 and O2, we can determine the limiting reactant. The limiting reactant is the reactant that is consumed completely in a reaction and determines the maximum amount of product that can be formed.

To figure out the limiting reactant, we have to compare the mole ratio between CH4 and O2 from the balanced equation:

From the balanced equation: CH4 + 2O2 -> CO2 + 2H2O

The mole ratio between CH4 and O2 is 1:2.

Since the ratio of moles of CH4 to O2 is less than the required 1:2 ratio, CH4 is the limiting reactant.

Now, let's calculate the moles of water produced from the limiting reactant:

moles of H2O = moles of CH4 x (moles of H2O / moles of CH4)
moles of H2O = 1.79 mol x (2 mol H2O / 1 mol CH4) ≈ 3.58 mol

Finally, we can convert the moles of water into grams using the molar mass of water (H2O):

molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

grams of H2O = moles of H2O x molar mass of H2O
grams of H2O = 3.58 mol x 18.02 g/mol ≈ 64.58 g

Therefore, when 28.7 grams of CH4 reacts with 45.8 grams of O2, approximately 64.58 grams of water can be formed.