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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.
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) 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?
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 Submit Request Answer
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 Submit Request Answer
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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