Solid sodium nitrate can be decomposed to solid sodium nitrite and
oxygen gas. If the percentage yield for the reaction is 80.4%, calculate the grams of sodium nitrite that will result from the decomposition of 955grams of sodium nitrate.
2 NaNO3 �¨ 2 NaNO2 + O2
If the reaction went to completion, the ratio of mass of NaNO2 product to NaNO3 reactant would be (23+14+32)/(23+14+48) = 69/85
The amount of NaNO2 formed would therefore be (69/85)(955) = 775 g
Take 80.4% of that.
607.02
To calculate the grams of sodium nitrite produced, we can start by finding the molar mass of sodium nitrate (NaNO3) and sodium nitrite (NaNO2).
The molar mass of NaNO3:
Na: 22.99 g/mol
N: 14.01 g/mol
O: 16.00 g/mol x 3 = 48.00 g/mol
Total molar mass = 22.99 + 14.01 + 48.00 = 84.00 g/mol
The molar mass of NaNO2:
Na: 22.99 g/mol
N: 14.01 g/mol
O: 16.00 g/mol x 2 = 32.00 g/mol
Total molar mass = 22.99 + 14.01 + 32.00 = 69.00 g/mol
Next, we can calculate the number of moles of sodium nitrate (NaNO3) using its given mass (955 grams) and its molar mass:
Moles of NaNO3 = mass / molar mass
Moles of NaNO3 = 955 g / 84.00 g/mol = 11.37 mol
According to the balanced equation, 2 moles of NaNO3 will produce 2 moles of NaNO2.
Therefore, the number of moles of NaNO2 produced will be the same as the number of moles of NaNO3:
Moles of NaNO2 = 11.37 mol
Finally, we can calculate the mass of sodium nitrite (NaNO2) produced using its molar mass:
Mass of NaNO2 = moles of NaNO2 x molar mass
Mass of NaNO2 = 11.37 mol x 69.00 g/mol = 784.53 grams
However, since the percentage yield of the reaction is given as 80.4%, we need to adjust the calculated mass accordingly:
Actual mass of NaNO2 = Mass of NaNO2 x (percentage yield / 100)
Actual mass of NaNO2 = 784.53 g x (80.4 / 100) = 631.56 grams
Therefore, the predicted grams of sodium nitrite produced from the decomposition of 955 grams of sodium nitrate is approximately 631.56 grams.
To solve this problem, we first need to determine the theoretical yield of sodium nitrite. The theoretical yield is the maximum amount of product that can be obtained from a given amount of reactant, assuming 100% efficiency.
The balanced chemical equation tells us that 2 moles of sodium nitrate (NaNO3) will produce 2 moles of sodium nitrite (NaNO2). So, the molar ratio between NaNO3 and NaNO2 is 2:2, which simplifies to 1:1.
1 mole of sodium nitrate has a molar mass of 85 grams (22.99 g/mol for sodium + 14.01 g/mol for nitrogen + (3 x 16.00 g/mol) for oxygen), so 955 grams of sodium nitrate is equivalent to:
(955 grams) / (85 grams/mol) = 11.24 moles of sodium nitrate
Since the molar ratio between NaNO3 and NaNO2 is 1:1, the number of moles of sodium nitrite produced will also be 11.24 moles.
Now, we can calculate the actual yield of sodium nitrite by multiplying the theoretical yield by the percentage yield:
Actual Yield = Theoretical Yield x Percentage Yield
Actual Yield = 11.24 moles x (80.4 / 100)
Finally, we need to convert moles of sodium nitrite to grams by using its molar mass. The molar mass of sodium nitrite (NaNO2) is 69 grams (22.99 g/mol for sodium + 14.01 g/mol for nitrogen + (2 x 16.00 g/mol) for oxygen):
Grams of Sodium Nitrite = Actual Yield x Molar Mass
Grams of Sodium Nitrite = (11.24 moles) x (69 grams/mol)
By performing the calculations:
Grams of Sodium Nitrite = 774.0 grams
Therefore, the decomposition of 955 grams of sodium nitrate will result in 774.0 grams of sodium nitrite, assuming an 80.4% yield.