Question

Balanced Equations:

Reaction 1: 1 NaHCO₃(s) → 1 NaOH(s) + 1 CO₂(g)
Reaction 2: 2 NaHCO₃(s) → 1 Na₂CO₃(s) + 1 CO₂(g) + 1 H₂O(g)
Reaction 3: 2 NaHCO₃(s) → 1 Na₂O(s) + 2 CO₂(g) + 1 H₂O(g)

Questions:

With 3.12 g of baking soda, how many moles of NaOH should be formed?

With 3.12 g of baking soda, how many moles of Na₂CO₃ should be formed?

With 3.12 g of baking soda, how many moles of Na₂O should be formed?

Now we can determine the mass for each.

With 3.12 g of baking soda, what mass of NaOH should be formed?

With 3.12 g of baking soda, what mass of Na₂CO₃ should be formed?

With 3.12 g of baking soda, what mass of Na₂O should be formed?

Answer 1

the balanced reactions are

1) NaHC03 ---> NaOH + C02

2) 2NaHC03 ---> Na2C03 + C02 + H20

3) 2 NaHC03 --> Na20 + 2C02 + H20

now

we know that

moles = mass / molar mass

so

moles of NaHC03 = 3.12 / 84 = 0.037143

now

a)

we can see that

moles of NaOH produced = moles of NaHC03 = 0.037143

now

mass = moles x molar mass

so

mass of NaOH = 0.037143 x 40 = 1.485

so

1.485 grams of NaOH is produced

b)

we can see that

moles of Na2c03 produced = 0.5 x moles of NaHC03 = 0.5 x 0.037143 = 0.0185715

now

mass = moles x molar mass

so

mass of Na2C03 = 0.0185715 x 106 = 1.97

so

1.97 grams of Na2C03 is produced

c)

we can see that

moles of Na20 produced = 0.5 x moles of NaHC03 = 0.5 x 0.037143 = 0.0185715

now

mass = moles x molar mass

so

mass of Na20 = 0.0185715 x 62 = 1.15

so

1.15 grams of Na2O is produced