Be sure to answer all parts. Calcium dihydrogen phosphate, Ca(H_2P0_4)_2, and so

Question

Be sure to answer all parts. Calcium dihydrogen phosphate, Ca(H_2P0_4)_2, and sodium hydrogen carbonate, NaHCO_3, are ingredients of baking powder that react with each other to produce CO_2, which causes dough or batter to rise: Ca(H_2PO_4)_2(s) + NaHCO_3(s) implies CO_2(g) + H_2O(g) + CaHPO_4(s) + Na_2HPO_4(s) [unbalanced] If the baking powder contains 31.0% NaHCO_3 and 35.0% Ca(H_2PO_4)_2 by mass: How many moles of CO_2 are produced from 7.73 g of baking powder? If 1 mol of CO_2 occupies 37.0 L at 350 degree Fahrenheit (a typical baking temperature), what volume of CO_2 is produced from 7.73 g of baking powder?

Explanation / Answer

the balanced equation is given by

3Ca(H2PO4)2 + 8NaHCO3 Ca3(PO4)2 + 4Na2HPO4 + 8CO2 + 8H2O

now

given

7.73 g baking powder

% mass = mass of component x 100 / total mass

so

mass of component = % mass x total mass / 100

so

mass of NaHC03 = 31 x 7.73 / 100 = 2.3963

mass of Ca(H2P04)2 = 35 x 7.73 / 100 = 2.7055

now

we know that

moles = mass / molar mass

so

moles of NaHC03 = 2.3963 / 84 = 0.02852738

moles of Ca(H2P04)2 = 2.7055 / 234 = 0.011561965

now

consider the given equation

3Ca(H2PO4)2 + 8NaHCO3 Ca3(PO4)2 + 4Na2HPO4 + 8CO2 + 8H2O

we can see that

moles of NaHC03 required = (8/3) x moles of Ca(H2PO4)2

so

moles of NaHC03 required = (8/3) x 0.011561965

moles of NaHC03 required = 0.030831908

but

only 0.02852738 moles of NaHC03 is present

so

NaHC03 is the limiting reagent

now

consider the given reaction

3Ca(H2PO4)2 + 8NaHCO3 Ca3(PO4)2 + 4Na2HPO4 + 8CO2 + 8H2O

we can see that

moles of C2 produced = moles of NaHC03 reacted

so

moles of C02 produced = 0.02852738

b)

now

volume = moles x 37

volume = 0.02852738 x 37

volume = 1.0555

so

1.0555 L of C02 is produced

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