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Given the data below, how many grams of NaCl would you expect to be formed in the reaction of excess HCl with the Na,CO3? The
You added HCI 1.00 mL at a time. When you added the last 1.00 mL amount that resulted in making bubbles, how many mL had you
How many moles of HCl were added to the reaction every time 1.00 mL of solution was added? (n = M x V) Choose the closest ans
2 When you added the last amount of HCl that bubbled, how many moles of HCI had you added to the solution? Choose the closest
What was the mass of the beaker plus NaCl after boiling off the water? Select one: a. 88.000 g O b.93.274 g c.87.206 g o d. 3
What was the mass of NaCl generated after boiling off the water? Select one: O a.0.936 g O b. 1.057 g O c.2.206 g O d. 0.226
The molecular weight of NaCl is 58.44 g/mol. Based on the number of grams of NaCl recovered, how many moles of NaCl were gene
Based on the number of moles of Na2CO3 consumed and NaCl produced, what is the stoichiometry of the reaction? Select one: O a
Why did the reaction stop bubbling after adding 7.00 mL of HCl solution? Select one: O a. All of the Na2CO3 had been reacted
What was the limiting reagent for the reaction? Select one: O a. H2O O b. Na2CO3 Ос. НСІ O d. Naci
Which of the following equations represents the chemical reaction? Select one: O a. 2 HCl(aq) + Na, CO3(aq) + NaCl(aq) + H2O(
Based on the balanced chemical equation, given that 2.000 g of Na2CO3 (molar mass = 105.989 g/mol) was added to the reaction,
Based on the balanced chemical equation, given that 2.000 g of NaCO3 (molar mass = 105.989 g/mol) was added to the reaction,
How does Na2CO3 act as a base? Select one: a. It acts as the conjugate acid. o b. The Coz binds to H from water, forming OH
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Answer #1

1. The balanced chemical equation for the reaction between Na2CO3 and HCl is written as

Na2CO3 + 2 HCl --------> 2 NaCl + CO2 + H2O

As per the stoichiometric equation,

1 mol Na2CO3 = 2 mols NaCl.

Gram molar mass of Na2CO3

= 105.989 g/mol.

Gram molar mass of NaCl = 58.443 g/mol.

Mass of Na2CO3 = 2.250 g.

Mol(s) Na2CO3 = (2.250 g)/(105.989 g/mol)

= 0.0212 mol.

Mol(s) NaCl produced = (0.0212 mol)*(2 mol NaCl)/(1 mol Na2CO3)

= 0.0424 mol.

Mass of NaCl produced = (0.0424 mol)*(58.443 g/mol)

= 2.4779 g

≈ 2.480 g

The closest answer is (a) 2.481 g (ans).

2. Need to know the molarity of HCl to answer the question.

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