Sodium carbonate reacts with hydrochloric acid as shown in an unbalanced chemical equation. What mass of Co2 is produced from the reaction of 2.94g Na2 Co2 with excess HCI?
Na2CO3(s)+HCI(aq)+ CO2(g)+H2O
A. 1.22 g
B. 2.44 g
C 2.94 g
D 5.88 g
E 7.08 g
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To determine the mass of CO2 produced, we need to use stoichiometry, which is the relationship between the coefficients in a balanced chemical equation.
The balanced equation is:
Na2CO3(s) + 2HCl(aq) -> CO2(g) + H2O + 2NaCl(aq)
From the balanced equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl to produce 1 mole of CO2.
First, we need to calculate the number of moles of Na2CO3 used:
moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
The molar mass of Na2CO3 is:
(2 x atomic mass of Na) + atomic mass of C + (3 x atomic mass of O)
= (2 x 22.99 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
= 105.99 g/mol
Using this molar mass, we find:
moles of Na2CO3 = 2.94 g / 105.99 g/mol
≈ 0.028 moles
Since the reaction is 1 mole of Na2CO3 reacts to produce 1 mole of CO2, the number of moles of CO2 produced is also 0.028 moles.
Finally, we can calculate the mass of CO2 produced:
mass of CO2 = moles of CO2 x molar mass of CO2
The molar mass of CO2 is:
atomic mass of C + (2 x atomic mass of O)
= 12.01 g/mol + (2 x 16.00 g/mol)
= 44.01 g/mol
Using this molar mass, we find:
mass of CO2 = 0.028 moles x 44.01 g/mol
≈ 1.23 g
Therefore, the answer is approximately 1.23 g.
A. 1.22 g would be the closest option.
To determine the mass of CO2 produced from the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl), we need to balance the chemical equation first.
The balanced chemical equation for the reaction is:
Na2CO3(s) + 2HCl(aq) → CO2(g) + 2NaCl(aq) + H2O(l)
From the balanced equation, we can see that 1 mole of Na2CO3 produces 1 mole of CO2.
To calculate the mass of CO2, we'll follow these steps:
Step 1: Convert the given mass of Na2CO3 to moles.
Molar mass of Na2CO3 = 22.99 g/mol (2 sodium atoms) + 12.01 g/mol (1 carbon atom) + (3 × 16.00 g/mol) (3 oxygen atoms) = 105.99 g/mol
Moles of Na2CO3 = mass of Na2CO3 / molar mass of Na2CO3
= 2.94 g / 105.99 g/mol
= 0.0277 mol
Step 2: Use the stoichiometry of the balanced equation to find the moles of CO2 produced.
From the balanced equation, the mole ratio between Na2CO3 and CO2 is 1:1.
Therefore, moles of CO2 = 0.0277 mol
Step 3: Convert moles of CO2 to mass.
Molar mass of CO2 (carbon dioxide) = 12.01 g/mol (1 carbon atom) + (2 × 16.00 g/mol) (2 oxygen atoms) = 44.01 g/mol
Mass of CO2 = moles of CO2 × molar mass of CO2
= 0.0277 mol × 44.01 g/mol
= 1.22 g
Therefore, the mass of CO2 produced from the reaction is 1.22 g.
The correct answer is A. 1.22 g.