Decomposition reaction help chemistry?

Question

What mass of NaHCO3 must decompose to produce 448 grams of Na2CO3 according to the equation 2 NaHCO3 ® CO2 H2O Na2CO3

Answer 1

First, work out the number of moles of Na2CO3 in 448 g of NaHCO3

moles = mass / molecular weight

molecular weight of Na2CO3 = (22.99 x 2) + 12.01 + (16.0 x 3) = 105.99 g/mol

therefore moles Na2CO3 = 448 g / 105.99 g/mol

= 4.23 moles of Na2CO3

2NaHCO3 -------------------> Na2CO3 + CO2 + H2O

The stoichiometry of the balanced equation tells you that 2 moles of NaHCO3 will decompose to give 1 mole of Na2CO3, therefore 4.23 moles of Na2CO3 would be produced from (2 x 4.23) moles of NaHCO3

= 8.46

So you must have started with 8.46 moles of NaHCO3

Now for mass of 8.46 moles of NaHCO3 we use again:

moles = mass / molecular weight

molecular weight of NaHCO3 = 22.99 + 1.008 + 12.01 + (16.0 x 3) = 84.008 g/mol

So

84.008 mol = mass / 84.008 g/mol

Therefore mass = 8.46 mol x 84.008 g/mol

= 710.71 g of NaHCO3

= 771 g (using correct sig. figures)