Sodium hydroxide reacts with carbon dioxide as follows: 2NaOH(s) + CO2(g) → NażCO3(s) + H2O(1) Part A Which reagent is the limiting reactant when 2.30 mol NaOH and 1.20 mol CO, are allowed to react? O N2OH O CO2 Submit Previous Answers v Correct The limiting reactant is the one that is completely consumed during the reaction. According to the reaction stoichiometry, two moles of NaOH are consumed per one mole of CO2 in a complete reaction. The amount of NAOH needed for a complete consumption of 1.20 mol CO2 is 2 mol NaOH mol NaOH needed to react 1.20 molCO x 1 molCO, 2.40 mol NaOH %3D Since only 2.30 mol NaOH are available, NaOH will run out before all CO2 reacts. Therefore, NaOH is the limiting reactant. Part B Part C How many moles of the excess reactant remain after the completion of the reaction?
Sodium hydroxide reacts with carbon dioxide as follows: 2NaOH(s) + CO2(g) → NażCO3(s) + H2O(1) Part A Which reagent is the limiting reactant when 2.30 mol NaOH and 1.20 mol CO, are allowed to react? O N2OH O CO2 Submit Previous Answers v Correct The limiting reactant is the one that is completely consumed during the reaction. According to the reaction stoichiometry, two moles of NaOH are consumed per one mole of CO2 in a complete reaction. The amount of NAOH needed for a complete consumption of 1.20 mol CO2 is 2 mol NaOH mol NaOH needed to react 1.20 molCO x 1 molCO, 2.40 mol NaOH %3D Since only 2.30 mol NaOH are available, NaOH will run out before all CO2 reacts. Therefore, NaOH is the limiting reactant. Part B Part C How many moles of the excess reactant remain after the completion of the reaction?
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Sodium hydroxide reacts with carbon dioxide as follows:
2NaOH(s)+CO2(g)→Na2CO3(s)+H2O(l)
How many moles of the excess reactant remain after the completion of the reaction?
Express your answer in moles to two decimal places.
Transcribed Image Text:I Review I Constants I Periodic Table
Sodium hydroxide reacts with carbon dioxide as follows:
2NAOH(s) + CO2 (g) → Na2 CO3 (s)+H2O(1)
Part A
Which reagent is the limiting reactant when 2.30 mol NaOH and 1.20 mol CO2 are allowed to react?
NaOH
CO2
Submit
Previous Answers
Correct
The limiting reactant is the one that is completely consumed during the reaction. According to the reaction stoichiometry, two
moles of NaOH are consumed per one mole of CO2 in a complete reaction. The amount of NaOH needed for a complete
consumption of 1.20 mol CO2 is
2 mol NaOH
mol NaOH needed to react
1.20 mol CO, x
1 mol-CO
2.40 mol NaOH
Since only 2.30 mol NaOH are available, NaOH will run out before all CO2 reacts. Therefore, NaOH is the limiting reactant.
Part B
Part C
How many moles of the excess reactant remain after the completion of the reaction?
Transcribed Image Text:I Review I Constants I Periodic Table
Sodium hydroxide reacts with carbon dioxide as follows:
2NAOH(s) + CO2(g) → Na2 CO3 (s) + H2O(1)
How many moles of Na2 CO3 can be produced?
Express your answer in moles to two decimal places.
n = 1.15 mol
Submit
Previous Answers
Correct
Since NaOH is the limiting reactant, the amount of available NaOH determines the amount of Na2 CO3 produced. Use the
number of moles of NaOH and the conversion factor derived from the stoichiometric relationship between NaOH and Na2 CO3
(2 mol NaOH:1 mol Na2 CO3) to calculate the number of moles of Na2 CO3 that can be made from 2.30 mol of NaOH:
1 mol Na2 CO3
mol Na2 CO3 produced
2.30 mol NaOH ×
2 mol NaOH
1.15 mol Na2 CO3
Part C
How many moles of the excess reactant remain after the completion of the reaction?
Express your answer in moles to two decimal places.
| ν ΑΣφ
?
n =
mol
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Expert Solution
Step 1
The provided chemical equation:
2NaOH (s)+CO2 (g)→Na2CO3 (s)+H2O (l)
2 mol 1 mol 1 mol 1 mol
Given
Moles of CO2=1.20 mol
Moles of NaOH=2.30 mol
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