Balanced Equations: Reaction 1: 1 NaOH(aq) + 1 HCl(aq) 1 NaCl(aq) + 1 H 2 O(l) R

Question

Balanced Equations:

Reaction 1: 1 NaOH(aq) + 1 HCl(aq) 1 NaCl(aq) +  1 H2O(l)
Reaction 2: 1 Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + 1 CO2(g) + 1 H2O(l)
Reaction 3: 1 Na2O(aq) +  2 HCl(aq) 2 NaCl(aq) + 1 H2O(l)

Questions:

Determine the number of moles of NaOH required to react with 0.01169 mol HCl.

Determine the number of moles of Na2CO3 required to react with 0.01169 mol HCl.

Determine the number of moles of Na2O required to react with 0.01169 mol HCl.

Now use the molar masses of the products to determine the mass of each solid required for the titration.

What mass of NaOH is required to react with 0.01169 mol HCl?

What mass of Na2O is required to react with 0.01169 mol HCl?

What mass of Na2CO3 is required to react with 0.01169 mol HCl?

Suppose you take 0.9600 g of the solid product. Now that you have done a few stoichiometric calculations we will ask directly for you to perform mass-mass calculations. Remember that along the way you will need to determine the number of moles and use the balanced equations above.

With 0.9600 g of NaOH and excess HCl, what mass of NaCl could be formed?

With 0.9600 g of Na2CO3 and excess HCl, what mass of NaCl could be formed?

With 0.9600 g of Na2O and excess HCl, what mass of NaCl could be formed?

Suppose you take 0.9600 g of the solid product. Now that you have done a few stoichiometric calculations we will ask directly for you to perform mass-mass calculations. Remember that along the way you will need to determine the number of moles and use the balanced equations above.

With 0.9600 g of NaOH and excess HCl, what mass of NaCl could be formed?

With 0.9600 g of Na2CO3 and excess HCl, what mass of NaCl could be formed?

With 0.9600 g of Na2O and excess HCl, what mass of NaCl could be formed?

Explanation / Answer

Answer – Given balanced reactions –

NaOH(aq) + 1 HCl(aq) 1 NaCl(aq) +  1 H2O(l)

Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + 1 CO2(g) + H2O(l)

Na2O(aq) +  2 HCl(aq) 2 NaCl(aq) + 1 H2O(l)

Moles of HCl = 0.01169 moles

Moles of NaOH

From the balanced equation –

1 moles of HCl = 1 mole of NaOH

So, 0.01169 moles of HCl = ?

= 0.01169 moles of NaOH

Moles of Na2CO3

From the balanced equation –

2 moles of HCl = 1 mole of Na2CO3

So, 0.01169 moles of HCl = ?

= 0.005845 moles of Na2CO3

Moles of Na2O

From the balanced equation –

2 moles of HCl = 1 mole of Na2O

So, 0.01169 moles of HCl = ?

= 0.005845 moles of Na2O

Mass of NaOH = 0.01169 moles * 40.0 g/mol

                          = 0.468 g of NaOH

Mass of Na2CO3 = 0.005845 moles * 105.98 g/mol

                            = 0.620 g of Na2CO3

Mass of Na2O = 0.005845 moles *61.98 g/mol

                        = 0.362 g of Na2O

Now mass of product

We are given the mass of limiting reactant, so we can convert the mass to moles and using the balanced reaction calculating the moles of product.

Moles of NaOH = 0.9600 g / 40.0 g.mol-1

                            = 0.024 moles

Moles of Na2CO3 = 0.9600 g / 105.98 g.mol-1

                             = 0.00906 moles

Moles of Na2O = 0.9600 g / 61.98 g.mol-1

                          = 0.0155 moles

Mass of NaCl from NaOH

From the balanced equation –

1 moles of NaOH = 1 mole of NaCl

So, 0.0240 moles of NaOH = ?

= 0.0240 moles NaCl

Mass of NaCl = 0.0240 moles *58.44 g/mol

                        = 1.40 g of NaCl

Mass of NaCl from Na2CO3

From the balanced equation –

1 moles of Na2CO3= 2 mole of NaCl

So, 0.00906 moles of Na2CO3= ?

= 0.0181 moles NaCl

Mass of NaCl = 0.0181 moles *58.44 g/mol

                        = 1.06 g of NaCl

Mass of NaCl from Na2O

From the balanced equation –

1 moles of Na2O= 2 mole of NaCl

So, 0.0155 moles of Na2O= ?

= 0.0310 moles NaCl

Mass of NaCl = 0.0310 moles *58.44 g/mol

                        = 1.81 g of NaCl

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