Sodium carbonate reacts with nitric acid according to the following equation:
Na2CO3+2HNO3-->2NaNO3+CO2+H2O
a. how many moles of Na2CO3 are required to produce 100.0 g of NaNO3?
b. if 7.50g of Na2CO3 reacts, how many moles of CO2 are produced?
if 7.50g of Na2CO3 reacts, how many moles of CO2 are produced?
a. To find the number of moles of Na2CO3 required to produce 100.0 g of NaNO3, we need to use the molar masses of Na2CO3 and NaNO3.
1. Calculate the molar mass of Na2CO3:
Na (sodium) has a molar mass of 22.99 g/mol
C (carbon) has a molar mass of 12.01 g/mol
O (oxygen) has a molar mass of 16.00 g/mol
So, the molar mass of Na2CO3 is:
(2 x 22.99 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol) = 105.99 g/mol
2. Calculate the number of moles of Na2CO3:
Number of moles = Mass / Molar mass
Number of moles = 100.0 g / 105.99 g/mol ≈ 0.943 mol
Therefore, approximately 0.943 moles of Na2CO3 are required to produce 100.0 g of NaNO3.
b. To find the number of moles of CO2 produced when 7.50 g of Na2CO3 reacts, we need to consider the stoichiometry of the balanced equation.
From the balanced equation, we see that the ratio between Na2CO3 and CO2 is 1:1. This means that for every 1 mole of Na2CO3, 1 mole of CO2 is produced.
1. Calculate the number of moles of Na2CO3:
Number of moles = Mass / Molar mass
Number of moles = 7.50 g / 105.99 g/mol ≈ 0.071 mol
2. Since the ratio between Na2CO3 and CO2 is 1:1, the number of moles of CO2 produced is also 0.071 mol.
Therefore, approximately 0.071 moles of CO2 are produced when 7.50 g of Na2CO3 reacts.
To solve these problems, we will first convert the given mass (in grams) to the number of moles using the molar mass of each compound. Then, we will use the stoichiometric coefficients from the balanced chemical equation to determine the desired quantities.
a. To find the number of moles of Na2CO3 required to produce 100.0 g of NaNO3, we need to use the molar masses of Na2CO3 and NaNO3. The molar mass of Na2CO3 (sodium carbonate) is 105.99 g/mol, and the molar mass of NaNO3 (sodium nitrate) is 85.00 g/mol.
First, let's calculate the number of moles of NaNO3 produced from 100.0 g of NaNO3:
Moles of NaNO3 = Mass of NaNO3 / Molar mass of NaNO3
Moles of NaNO3 = 100.0 g / 85.00 g/mol ≈ 1.176 moles
According to the stoichiometry of the balanced equation, 2 moles of NaNO3 are produced from 1 mole of Na2CO3. So, the number of moles of Na2CO3 required can be calculated as follows:
Moles of Na2CO3 = Moles of NaNO3 / 2
Moles of Na2CO3 = 1.176 moles / 2 ≈ 0.588 moles
Therefore, approximately 0.588 moles of Na2CO3 are required to produce 100.0 g of NaNO3.
b. To find the number of moles of CO2 produced from 7.50 g of Na2CO3, we need to use the molar mass of CO2, which is 44.01 g/mol.
First, let's calculate the number of moles of Na2CO3:
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
Moles of Na2CO3 = 7.50 g / 105.99 g/mol ≈ 0.071 moles
According to the stoichiometry of the balanced equation, 1 mole of Na2CO3 reacts to produce 1 mole of CO2. So, the number of moles of CO2 produced is equal to the number of moles of Na2CO3:
Moles of CO2 = Moles of Na2CO3 ≈ 0.071 moles
Therefore, approximately 0.071 moles of CO2 are produced when 7.50 g of Na2CO3 react.
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html