Sodium nitrite is prepared by a reaction of sodium nitrate and carbon: NaNO3 + C → NaNO2 + CO2 Calculate the masses of reactants needed for the preparation of 10 g of NaNO2. Carbon is used in 10% excess. The atomic/molecular masses are: NaNO2= 69.0 g/mol , NaNO3=85.0 g/mol and C=12 g/mol
Keep in mind that the law of conservation mass. That's All.
(Sorry if I seem like a prick)
To calculate the masses of the reactants needed for the preparation of 10 g of NaNO2, we need to use the stoichiometry of the balanced equation and consider the percentage excess of carbon.
First, let's identify the balanced equation and the stoichiometry:
NaNO3 + C → NaNO2 + CO2
From the balanced equation, we can see that 1 mole of NaNO3 reacts with 1 mole of C to produce 1 mole of NaNO2 and 1 mole of CO2.
Next, let's calculate the moles of NaNO2:
Moles of NaNO2 = mass of NaNO2 / molar mass of NaNO2
= 10 g / 69.0 g/mol
= 0.1449 mol (approximately)
Now, let's calculate the moles of NaNO3:
Since the stoichiometry of NaNO2 to NaNO3 is 1:1, the moles of NaNO3 will be equal to the moles of NaNO2:
Moles of NaNO3 = 0.1449 mol
To calculate the moles of C, we need to consider the 10% excess:
Moles of C = Moles of NaNO3 * (1 + 10%)
= 0.1449 mol * 1.10
= 0.1594 mol (approximately)
Finally, let's calculate the masses of the reactants:
Mass of NaNO3 = Moles of NaNO3 * molar mass of NaNO3
= 0.1449 mol * 85.0 g/mol
= 12.32 g (approximately)
Mass of C = Moles of C * molar mass of C
= 0.1594 mol * 12 g/mol
= 1.91 g (approximately)
Therefore, to prepare 10 g of NaNO2, you would need approximately 12.32 g of NaNO3 and 1.91 g of C (with a 10% excess of carbon).