What mass, in grams, of sodium carbonate is required for complete reaction with 8.35g of nitric acid to produce sodium nitrate, carbon dioxide, and water? [Show work]
2HNO3 + Na2CO3 ==> CO2 + H2O + 2NaNO3
mols HNO3 = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols HNO3 to mols Na2CO3.
Then convert mols Na2CO3 to grams. g = mols x molar mass = ?
To find the mass of sodium carbonate required for the reaction, we need to use stoichiometry, which is the numerical relationship between the amounts of reactants and products in a balanced chemical equation.
First, let's write the balanced equation for the reaction between sodium carbonate (Na2CO3) and nitric acid (HNO3):
2 Na2CO3 + 2 HNO3 → 2 NaNO3 + CO2 + H2O
From the balanced equation, we can see that 2 moles of sodium carbonate react with 2 moles of nitric acid to produce 2 moles of sodium nitrate, 1 mole of carbon dioxide, and 1 mole of water.
To find the molar mass of sodium carbonate (Na2CO3), we add the molar masses of its individual elements:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Molar mass of Na2CO3 = (2 × Na) + C + (3 × O) = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
Now we can set up a conversion factor to determine the mass of sodium carbonate required:
1 mol Na2CO3 = 105.99 g Na2CO3
To find the number of moles of nitric acid (HNO3) present, we use its molar mass:
H: 1.01 g/mol
N: 14.01 g/mol
O: 16.00 g/mol
Molar mass of HNO3 = (1 × H) + N + (3 × O) = 1.01 + 14.01 + (3 × 16.00) = 63.01 g/mol
8.35 g HNO3 × (1 mol HNO3 / 63.01 g HNO3) = 0.1324 mol HNO3
According to the balanced equation, 2 moles of Na2CO3 react with 2 moles of HNO3. Therefore, the number of moles of Na2CO3 required is also 0.1324 mol.
Finally, we can calculate the mass of sodium carbonate required:
0.1324 mol Na2CO3 × (105.99 g Na2CO3 / 1 mol Na2CO3) = 14.00 g Na2CO3
Therefore, 14.00 grams of sodium carbonate is required for complete reaction with 8.35 grams of nitric acid to produce sodium nitrate, carbon dioxide, and water.
To determine the mass of sodium carbonate required for the reaction, we need to use the balanced chemical equation:
2 NaHCO3 + HNO3 -> NaNO3 + CO2 + H2O
From the equation, we can see that the mole ratio between sodium carbonate (NaHCO3) and nitric acid (HNO3) is 2:1.
1. Calculate the number of moles of nitric acid:
Given mass of nitric acid: 8.35g
Molar mass of nitric acid (HNO3): 63.01 g/mol
Number of moles of nitric acid = given mass / molar mass
Number of moles of nitric acid = 8.35g / 63.01 g/mol
Number of moles of nitric acid ≈ 0.132 moles
2. Since the mole ratio between sodium carbonate and nitric acid is 2:1, we need twice the number of moles of sodium carbonate compared to nitric acid.
Number of moles of sodium carbonate = 2 * number of moles of nitric acid
Number of moles of sodium carbonate ≈ 2 * 0.132 moles
Number of moles of sodium carbonate ≈ 0.264 moles
3. Finally, calculate the mass of sodium carbonate:
Molar mass of sodium carbonate (NaHCO3): 84.01 g/mol
Mass of sodium carbonate = number of moles * molar mass
Mass of sodium carbonate ≈ 0.264 moles * 84.01 g/mol
Mass of sodium carbonate ≈ 22.16g
Therefore, approximately 22.16 grams of sodium carbonate is required for complete reaction with 8.35 grams of nitric acid.