In: Chemistry
Be sure to answer all parts. Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise: Ca(H2PO4)2(s) + NaHCO3(s) → CO2(g) + H2O(g) + CaHPO4(s) + Na2HPO4(s)[unbalanced] If the baking powder contains 31.0% NaHCO3 and 35.0% Ca(H2PO4)2 by mass: (a) How many moles of CO2 are produced from 9.66 g of baking powder? mol CO2 (b) If 1 mol of CO2 occupies 37.0 L at 350°F (a typical baking temperature), what volume of CO2 is produced from 9.66 g of baking powder?
Answer – We are given, reaction –
Ca(H2PO4)2(s) + 2 NaHCO3(s) ----> 2 CO2(g) + 2 H2O(g) + CaHPO4(s) + Na2HPO4(s)
Mass percent of NaHCO3(s) = 31.0% , mass % of Ca(H2PO4)2(s) = 35.0 %
a) Mass of baking powder = 9.66 g , moles of CO2 =
now we need to calculate the mass of each reactant
mass of NaHCO3(s) = 31.0%*9.66 g / 100 %
= 2.995 g
mass of Ca(H2PO4)2(s) = 35.0 % 9.66 g / 100 %
= 3.381 g
Moles of NaHCO3(s) = 2.995 g / 84.00 g.mol-1
= 0.0356 moles
Moles of Ca(H2PO4)2(s) = 3.381 g / 234.05 g.mol-1
= 0.0144 moles
Now we need to calculate the limiting reactant
Moles of CO2 from Ca(H2PO4)2(s)
We know
1 moles of Ca(H2PO4)2(s) = 2 moles of CO2
So, 0.0144 moles of Ca(H2PO4)2(s) = ?
= 0.0288 moles of CO2
Moles of CO2 from the NaHCO3(s)
1 moles of NaHCO3(s) = 2 moles of CO2
So, 0.0356 moles of NaHCO3(s) = ?
= 0.0356 moles of CO2
So, the lowest moles of the CO2 from the Ca(H2PO4)2(s) , so limiting eactant is Ca(H2PO4)2(s) .
Moles of CO2 = 0.0288 moles
So, 0.0288 moles of CO2 are produced from 9.66 g of baking powder.
b) We are given, 1 moles of CO2 = 37.0 L at 350°F , mass of baking powder = 9.66 g
We already calculated moles of CO2 produced from 9.66 g of baking powder
So,
1 moles of CO2 = 37.0 L
0.0288 moles of CO2 = ?
= 1.07 L
So, 1.07 L volume of CO2 is produced from 9.66 g of baking powder