Question

stoichiometry of an acid basRecord the following masses. a mass of empty beaker (g) 85.000g b mass of beaker plus Na2CO3 (g) 87.000g c mass of Na2CO3 (g) 2.000g Data Analysis Convert the mass of Na2CO3 to moles, given its molar mass (MM) of 105.989 g/mol. 2g/105.989gmol = 0.018869=0.01887

Experiment 2: Neutralization Reaction Lab Results What was the acid in the reaction? HCL Data Analysis How many moles of HCl did you add to fully neutralize the Na2CO3 solution?

Experiment 3: Isolate Sodium Chloride Lab Results Record the following masses: a mass of empty beaker (g) 85.000g b madd of beaker plus NaCl (after boiling off the water) (g) 87.206g Data Analysis

Convert the mass of NaCl to moles, given its molar mass (MM) of 58.443 g/mol. Conclusions Calculate the experimentally determined molar ratio of Na2CO3 to NaCl using the formula below. molar ratio = (mol Na2CO3) / (mol NaCl)

Use the theoretical molar ratio to calculate the theoretical yield of NaCl in grams from 2.000 g of Na2CO3? The percent yield is the ratio of the actual amount of a product to the theoretical amount.

Calculate the percent yield of NaCl as shown below. %yield = (experimental yield) / (theoretical yield) × 100

Given the data below, how many grams of CO2 would you expect to be formed in the reaction of excess HCl with the Na2CO3? mass of empty beaker (g) 84.000 mass of beaker plus Na2CO3 (g) 86.375 mass of Na2CO3 (g) 2.375 e reaction

Please show work

Answer 1

Answer - We are given, mass of Na₂CO₃ = 2.00 g

Moles of Na₂CO₃ = 2.00 g / 105.988 g.mol^-1

                              = 0.0189 moles

Number of moles of HCl fully neutralize the Na₂CO₃ solution-

We know the reaction –

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

From the balanced reaction –

1 moles of Na₂CO₃ = 2 moles of HCl

So, 0.0189 moles Na₂CO₃ = ?

= 0.0377 moles of HCl

Mass of NaCl = mass of beaker plus NaCl - mass of empty beaker

                       = 87.206 g – 85.000 g

                      = 2.206 g

Moles of NaCl = 2.206 g / 58.443 g.mol^-1

                         = 0.0377 moles

Theoretical yield of NaCl

From the balanced reaction –

1 moles of Na₂CO₃ = 2 moles of NaCl

So, 0.0189 moles Na₂CO₃ = ?

= 0.0377 moles of NaCl

Theoretical yield of NaCl = 0.0377 moles of NaCl *58.443 g/mol

                                          = 2.206 g

We know,

Percent yield = (experimental yield) / (theoretical yield) × 100 %

                      = 2.206 g / 2.206 g * 100 %

                      = 100 %

Mass of CO₂ –

Mass of Na₂CO₃ = 86.375 g – 84.000 g

                            = 2.375 g

Moles of Na₂CO₃ = 2.375 g / 105.988 g.mol^-1

                              = 0.0224 moles

From the balanced reaction –

1 moles of Na₂CO₃ = 1 moles of CO₂

So, 0.0224 moles Na₂CO₃ = ?

= 0.0224 moles of CO₂

Mass of CO₂ = 0.0224 moles of CO₂ * 44.0 g/mol

                     = 0.986 g of CO₂