These are the answers I got for the following problems - am I right?
1) How many grams of water, H2O, and propene, C3H6, can be formed from the reaction of 6.0 g of 2-propanol, C3H8O?
C3H8O ----> C3H6 + H2O
1.8g H2O and 4.2g C3H6
2) How many moles of water, H2O, can be produced from the reaction of 16 g of methane, CH4, with an unlimited supply of oxygen, O2? How many grams of water is this? The reaction is:
CH4 + 2O2 ----> CO2 + 2H2O
36g water, which is two moles.
Those look ok to me. Good work.
Let's check your answers step by step:
1) To determine the correct mass of water (H2O) and propene (C3H6) formed, we need to use the given molar ratios. The balanced chemical equation shows that for every 1 mole of 2-propanol (C3H8O), we get 1 mole of propene (C3H6) and 1 mole of water (H2O).
First, we need to convert the given mass of 2-propanol to moles using its molar mass. The molar mass of 2-propanol (C3H8O) is:
(3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (1 * 16.00 g/mol) = 60.11 g/mol
Now, using the given mass of 6.0 g, we can calculate the number of moles:
moles of 2-propanol = mass / molar mass = 6.0 g / 60.11 g/mol ≈ 0.0999 mol
Since the molar ratio is 1:1:1 for 2-propanol to propene to water, we can conclude that:
moles of water = 0.0999 mol
moles of propene = 0.0999 mol
Finally, we can convert the moles of water and propene back to grams using their molar masses:
mass of water = moles of water * molar mass of water = 0.0999 mol * 18.015 g/mol ≈ 1.801 g
mass of propene = moles of propene * molar mass of propene = 0.0999 mol * 42.08 g/mol ≈ 4.209 g
Therefore, the correct mass of water formed is approximately 1.801 grams and the correct mass of propene formed is approximately 4.209 grams. Your answer of 1.8g H2O and 4.2g C3H6 is correct.
2) To determine the moles of water (H2O) produced, we need to use the given mass of methane (CH4) and its molar mass. The molar mass of methane is:
(1 * 12.01 g/mol) + (4 * 1.01 g/mol) = 16.05 g/mol
Now, using the given mass of 16 g, we can calculate the number of moles:
moles of methane = mass / molar mass = 16 g / 16.05 g/mol ≈ 0.996 mol
According to the balanced chemical equation, 1 mole of methane produces 2 moles of water. So, the moles of water produced will be:
moles of water = 2 * moles of methane = 2 * 0.996 mol = 1.992 mol
To determine the mass of water produced, we can use the molar mass of water:
mass of water = moles of water * molar mass of water = 1.992 mol * 18.015 g/mol ≈ 35.9 g
Therefore, the correct number of moles of water produced is approximately 1.992 moles, and the correct mass of water produced is approximately 35.9 grams. Your answer of 36g water, which is two moles, is correct.
To check if the answers you got are correct, let's go through the calculations and verify them.
1) To find the grams of water (H2O) and propene (C3H6) formed from 6.0 g of 2-propanol (C3H8O), we need to determine the stoichiometry of the reaction. The balanced equation tells us that 1 molecule of 2-propanol (C3H8O) produces 1 molecule of propene (C3H6) and 1 molecule of water (H2O).
First, let's find the molar mass of C3H8O:
C = 12.01 g/mol x 3 = 36.03 g/mol
H = 1.01 g/mol x 8 = 8.08 g/mol
O = 16.00 g/mol x 1 = 16.00 g/mol
Total molar mass of C3H8O = 36.03 + 8.08 + 16.00 = 60.11 g/mol
Now, use the molar mass and the given mass of 2-propanol to convert grams to moles:
moles of 2-propanol = 6.0 g / 60.11 g/mol ≈ 0.10 mol
Since the reaction stoichiometry tells us that the mole ratio between C3H8O and H2O is 1:1, the same number of moles of water will be produced.
moles of H2O = 0.10 mol
Now, convert moles back to grams using the molar mass of water:
grams of H2O = moles of H2O x molar mass = 0.10 mol x 18.02 g/mol = 1.80 g of H2O
Therefore, the correct answer is 1.80 g of H2O, which matches your calculated value.
Next, let's calculate the grams of propene (C3H6) formed:
moles of C3H6 = 0.10 mol
Now, convert moles back to grams using the molar mass of propene:
grams of C3H6 = moles of C3H6 x molar mass = 0.10 mol x (3 x 12.01 g/mol + 6 x 1.01 g/mol) = 4.20 g of C3H6
Therefore, the correct answer is 4.20 g of C3H6, which also matches your calculated value.
2) To find the moles of water (H2O) produced from the reaction of 16 g of methane (CH4), we need to note the stoichiometry of the reaction. The balanced equation tells us that 1 molecule of methane (CH4) produces 2 molecules of water (H2O) and 1 molecule of carbon dioxide (CO2).
First, find the molar mass of CH4:
C = 12.01 g/mol
H = 1.01 g/mol x 4 = 4.04 g/mol
Total molar mass of CH4 = 12.01 + 4.04 = 16.05 g/mol
Now, use the molar mass and the given mass of methane to convert grams to moles:
moles of CH4 = 16 g / 16.05 g/mol ≈ 1.00 mol
Since the reaction stoichiometry tells us that the mole ratio between CH4 and H2O is 1:2, twice the number of moles of water will be produced.
moles of H2O = 2 x moles of CH4 = 2 x 1.00 mol = 2.00 mol
Now, convert moles back to grams using the molar mass of water:
grams of H2O = moles of H2O x molar mass = 2.00 mol x 18.02 g/mol = 36.04 g of H2O
Therefore, the correct answer is 36.04 g of H2O, which matches your calculated value. Also, since 2 moles of H2O weighs 36.04 g, your answer is correct in stating that it is two moles of water.
To summarize, the answers you provided are both correct:
1) The reaction of 6.0 g of 2-propanol will produce 1.8 g of H2O and 4.2 g of C3H6.
2) The reaction of 16 g of methane will produce 36 g of H2O, which is equivalent to two moles of water.