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The equations are 2NaHCO3 -> Na2CO3 + CO2 + H20, NaHCO3 -> NaOH + CO2, 2NaHCO3 -> Na2O + 2CO2 + H20
5.0 grams of baking soda is heated in a metal crucible for 10 minutes. The leftover product is 2.92 grams.
I've already done the math and 5.0 grams of baking soda is .06 mol of baking soda
Each of the equations calculated is left with 2.4g of NaOH, 1.86 g Na2O, and 3.18g Na2Co3.
How can I determine which of these is the product that was made, and how can I determine the percent yield of the product

Respuesta :

Answer:

The product made is Na2CO3.

The % yield is 91.8 %

Explanation:

Step 1: Data given

Mass of baking soda (NaHCO3) = 5.0 grams

Molar mass of NaHCO3 = 84.0 g/mol

Each of the equations calculated is left with 2.4g of NaOH, 1.86 g Na2O, and 3.18g Na2Co3.

Step 2: The balanced equations

2NaHCO3 → Na2CO3 + CO2 + H20

NaHCO3 → NaOH + CO2

2NaHCO3 → Na2O + 2CO2 + H20

Step 3: Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

Moles NaHCO3 = 5.0 grams / 84.0 g/mol

Moles NaHCO3 = 0.060 moles

Step 4: Calculate moles of products

For 2 moles NaHCO3 we'll have 1 mol Na2CO3

For 0.060 moles NaHCO3 we'lll have 0.060 / 2 = 0.030 moles Na2CO3

For  1 mol NaHCO3 we'll have 1 mol NaOH

For 0.060 moles NaHCO3 we'll have 0.060 moles NaOH

For 2 moles NaHCO3 we'll have 1 mol Na2O

For 0.060 moles NaHCO3 we'll have 0.030 moles Na2O

Step 5: Calculate mass of products

Mass = moles * molar mass

Mass of Na2CO3 = 0.030 moles * 105.99 g/mol = 3.18 grams

Mass of NaOH = 0.060 moles * 40.0 g/mol = 2.4 grams

Mass of Na2O = 0.030 moles *61.98 g/mol = 1.86 grams

Step 6: Calculate the percent yield

% yield = actual yield / theoretical yield

% yield Na2CO3 = (2.92 grams / 3.18 grams) *100% =  91.8 %

% yield NaOH = (2.92 grams / 2.4 grams ) *100% = 121.6 %

% yield of Na2O = (2.92 grams / 1.86 grams ) * 100% = 157 %

The product made is Na2CO3, the other reactions have a % yield greater than 100 %

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