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Blababa [14]
3 years ago
10

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45

g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon)
Chemistry
1 answer:
musickatia [10]3 years ago
5 0

Answer:

For 1 antacid tablet (in ballon1) we get .0173 moles of CO2

for 2 tablets (in balloon 2) we get: 2*0,0173=  0.0346 moles of CO2

For 3 tablets (in balloon 3) we get 3* 0.0173 = 0.0519 moles of CO2

Explanation:

The complete question:

Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon).

Balloon 1 had 1 antacid tab

Baloon 2 had 2

Balloon 3 had 3

Step 1: Data given

1.45 g of NaHCO3 is present in an antacid tablet

Molar mass of NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

NaHCO3 + H2O → NaOH + H2O + CO2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

1.45g / 84.0 g/mol = .0173 moles

Step 4: Calculate moles CO2

For 1 mol NaHCO3 we need 1 mol H2O to produce 1 mol NaOH 1 mol H2O and 1 mol CO2

For 0.0173 moles NaHCO3 we'll get 0.0173 moles CO2

so for 1 antacid tablet we get .0173 moles of CO2

for 2 tablets we get: 2*0,0173 =  0.0346 moles of CO2

For 3 tablets we get 3* 0.0173 = 0.0519 moles of CO2

 

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Answer: Mg_{1}Cl_{2}O_{8}

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

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Step 1 : convert given masses into moles.

Moles of Mg=\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles

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Moles of O =\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = \frac {0.45}{0.45}=1

For Cl = \frac {0.89}{0.45}=2

For O= \frac {3.58}{0.45}=8

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is Mg_{1}Cl_{2}O_{8}

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