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.
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
3 answers
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 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).