(I corrected it )
1. Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liters of hydrogen gas are formed at STP?
2Na+2H2O—>2NaOH+H2
My work: For #1 Na would have 2 moles and H would have 1? Make 31.5g x 2/1=63? Or 31.5g divided by 2 x 1= 15.7?
(If im wrong can you help me)
2. Sodium reacts with oxygen to produce sodium oxide as described by the balanced equation below. If 54.1g of sodium reacts with excess oxygen gas to produce 61.8g Of sodium oxide, what is the percent yield? (Hint: be sure to calculate theoretical yield first)
4Na+O2–>2Na2O
I’m having trouble solving #2 can someone give me a format on how to solve for the theoretical yield.
I know how to solve the problem just not the theoretical yield that’s the part I’m struggling on.
I’m not sure but I think the equation is already balanced. (If not can someone tell me so I can correctly try to fix it)
These questions both come from an online school chemistry test. This student is violating our honor code. Please stop helping students cheat by simply giving them the answers.
For problem #1, you are correct that a mole ratio of 2:1 is appropriate based on the balanced equation. However, your calculation is not accurate. Here is the correct step-by-step calculation:
1. Convert the given mass of sodium (31.5g) to moles using the molar mass of sodium (22.99 g/mol):
31.5g Na x (1 mol Na / 22.99 g Na) = 1.37 mol Na
2. Use the mole ratio from the balanced equation (2 moles of Na react to produce 1 mole of H2) to determine the moles of H2 gas formed:
1.37 mol Na x (1 mol H2 / 2 mol Na) = 0.685 mol H2
3. Calculate the volume of hydrogen gas at STP (standard temperature and pressure). At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore:
0.685 mol H2 x (22.4 L / 1 mol) = 15.34 L H2 (rounded to two decimal places)
So, when 31.5g of sodium reacts with excess water, approximately 15.34 liters of hydrogen gas are formed at STP.
Now, moving on to problem #2:
The given equation is already correctly balanced:
4 Na + O2 → 2 Na2O
To calculate the theoretical yield and percent yield, follow these steps:
1. Determine the moles of sodium oxide produced by dividing the given mass (61.8g Na2O) by the molar mass of Na2O (61.98 g/mol):
61.8g Na2O x (1 mol Na2O / 61.98 g Na2O) = 1.00 mol Na2O
2. Use the mole ratio from the balanced equation (4 moles of Na react to produce 2 moles of Na2O) to calculate the theoretical moles of Na:
1.00 mol Na2O x (4 mol Na / 2 mol Na2O) = 2.00 mol Na
3. Convert the theoretical moles of Na to grams using the molar mass of Na (22.99 g/mol):
2.00 mol Na x (22.99 g Na / 1 mol Na) = 45.98g Na (rounded to two decimal places)
4. Calculate the percent yield by dividing the given mass (54.1g Na) by the theoretical yield (45.98g Na) and multiplying by 100:
percent yield = (54.1g Na / 45.98g Na) x 100 = 117.76% (rounded to two decimal places)
The percent yield is 117.76%, which indicates that the actual yield (61.8g Na2O) is greater than the theoretical yield (45.98g Na), which is not physically possible. Therefore, the percent yield should be capped at 100%.
To solve for the theoretical yield in question 2, you first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed.
1. Calculate the moles of sodium (Na):
- Given mass of sodium = 54.1g
- Calculate the moles of sodium using the molar mass of sodium (22.99g/mol): 54.1g / 22.99g/mol = 2.35 mol
2. Calculate the moles of sodium oxide (Na2O):
- Given mass of sodium oxide = 61.8g
- Calculate the moles of sodium oxide using the molar mass of sodium oxide (61.97g/mol): 61.8g / 61.97g/mol = 0.998 mol
3. Determine the stoichiometric ratio between sodium and sodium oxide:
- From the balanced equation: 4Na + O2 -> 2Na2O
- The ratio between sodium and sodium oxide is 4:2 or 2:1 (2 moles of sodium oxide are formed for every 1 mole of sodium)
4. Calculate the theoretical yield:
- Since the ratio is 2:1, we can say that 2 moles of Na2O are formed for every 1 mole of Na.
- The moles of sodium oxide formed = 1 mole Na x (2 moles Na2O / 1 mole Na) = 2 moles Na2O
5. Convert moles of sodium oxide to grams using the molar mass:
- The molar mass of sodium oxide is 61.97g/mol
- The theoretical mass of sodium oxide = 2 moles Na2O x 61.97g/mol = 123.94g
Now that you have the theoretical yield (123.94g), you can calculate the percent yield by dividing the given mass of sodium oxide by the theoretical yield, and then multiplying by 100:
Percent yield = (given mass of sodium oxide / theoretical yield) x 100
I hope this helps!
1. molar mass of Na is 23.0 g
moles of Na = 31.5 / 23.0
moles of H2 is half of Na
... from reaction equation
a mole of ANY gas is 22.4 L at STP
2. the equation is balanced
theoretical is what you get if ALL the sodium reacts with oxygen to form sodium oxide
54.1/23.0 moles of Na, theoretically yields 54.1/46.0 moles of Na2O
... molar mass of Na2O is 62.0 g
find the theoretical product mass and compare it to the actual to get the % yield