What is the limiting reactant/excess reactant and maximum theoretical product in grams if given starting reactants of
5.0g Na() and 5.0 g O21) using the balanced equation 4Na(g) + 0218) -+ 2NazOl»
A. Limiting reactant:
Na,
B. Limiting reactant:
02,
C. Limiting reactant:
Na,
D. Limiting reactant :
02,
E. Limiting reactant : Na,
F. Limiting reactant :
02,
Excess Reactant: Oz
M.T.P. = 470. grams Na20
Excess Reactant: Na, M.T.P. = 480. grams NazO
Excess Reactant: O2, M.T.P. = 43. grams Naz0
Excess Reactant: Na, M.T.P. = 477. grams Na20
Excess Reactant: 02, M.T.P. = 6.7. grams NazO
Excess Reactant: Na, M.T.P. = 19.3. grams Na20
The correct answer is A. Limiting reactant: Na, Excess Reactant: O2, M.T.P. = 470. grams Na2O.
To determine the limiting reactant, we need to calculate the amount of product that each reactant can form based on the balanced equation.
For Na: 5.0 g Na x (1 mol Na/22.99 g Na) x (2 mol Na2O/4 mol Na) x (62.00 g Na2O/1 mol Na2O) = 32.1 g Na2O
For O2: 5.0 g O2 x (1 mol O2/32.00 g O2) x (2 mol Na2O/1 mol O2) x (62.00 g Na2O/1 mol Na2O) = 77.2 g Na2O
We can see that Na can only form 32.1 g of Na2O, while O2 can form 77.2 g of Na2O. This means that Na is the limiting reactant, and O2 is in excess.
To determine the maximum theoretical product, we use the amount of limiting reactant available:
5.0 g Na x (1 mol Na/22.99 g Na) x (2 mol Na2O/4 mol Na) x (62.00 g Na2O/1 mol Na2O) = 14.3 g Na2O
However, we need to divide this by 2 because the balanced equation shows that 2 moles of Na2O are formed for every 4 moles of Na.
So the maximum theoretical product is 14.3 g Na2O/2 = 7.15 g Na2O.
However, we have to be careful because we were given the amount of O2 to be 5.0 g. This means that we have excess O2 left over after the reaction is complete.
To find the actual amount of Na2O formed, we need to use the amount of limiting reactant available (5.0 g Na) and the stoichiometry of the balanced equation:
5.0 g Na x (1 mol Na/22.99 g Na) x (2 mol Na2O/4 mol Na) x (62.00 g Na2O/1 mol Na2O) = 14.3 g Na2O
This is the same as the maximum theoretical product we calculated earlier, so the actual amount of Na2O formed is 14.3 g.
Therefore, the correct answer is A. Limiting reactant: Na, Excess Reactant: O2, M.T.P. = 470. grams Na2O.
Determine the answer to the following equation with correct number of significant figures 13.96 - 4.9102+71.5=
The answer is 80.0898.
To determine the correct number of significant figures, we look at the least precise measurement, which in this case is 4.9102. It has 5 significant figures.
When performing addition or subtraction, we need to round the answer to the same number of decimal places as the least precise measurement.
Therefore, the final answer is 80.090 (rounded to one decimal place).
To determine the limiting reactant and the maximum theoretical product in this reaction, we need to calculate the amount of product that can be formed using each reactant, and then compare the results.
Step 1: Convert the given masses of Na and O2 to moles.
- Na:
There is a molar mass of Na = 22.99 g/mol.
Convert 5.0 g Na to moles using the formula:
moles of Na = mass of Na / molar mass of Na
- O2:
There is a molar mass of O2 = 32.00 g/mol.
Convert 5.0 g O2 to moles using the formula:
moles of O2 = mass of O2 / molar mass of O2
Step 2: Use the stoichiometry of the balanced equation to find the moles of the product that can be formed from each reactant.
From the balanced equation: 4Na(g) + O2(g) -> 2Na2O(l)
- Na:
Since the coefficient of Na in the equation is 4, we can use the mole ratio to find the moles of Na2O produced from Na:
moles of Na2O = moles of Na / 4 (according to the stoichiometric ratio)
- O2:
Since the coefficient of O2 in the equation is 1, we can use the mole ratio to find the moles of Na2O produced from O2:
moles of Na2O = moles of O2 / 1 (according to the stoichiometric ratio)
Step 3: Compare the moles of Na2O produced from each reactant.
Whichever reactant produces the smaller amount of the product is the limiting reactant. The reactant that produces the larger amount is the excess reactant.
Step 4: Convert the moles of Na2O to grams of Na2O to find the maximum theoretical product.
- Molar mass of Na2O = 61.98 g/mol
Use the formula:
mass of Na2O = moles of Na2O * molar mass of Na2O
Step 5: Determine the correct answer.
Based on the calculations, the correct answers are:
- Limiting Reactant: Na
- Excess Reactant: O2
- Maximum Theoretical Product: 19.3 grams Na2O
Therefore, the correct answer is:
Excess Reactant: Na, M.T.P. = 19.3 grams Na2O.