8.4g of sodium hydrogen carbonate is completely decomposed by heat.Calculate:

Question

a)the mass of residue produced
b)Volume of carbon(IV)oxide produced at stp

Answer 1

To calculate the mass of residue produced, we need to determine the chemical reaction that occurs when sodium hydrogen carbonate (NaHCO3) decomposes.

The balanced chemical equation for the reaction is:
2 NaHCO3 → Na2CO3 + CO2 + H2O

From the equation, we can see that 2 moles of NaHCO3 produce 1 mole of Na2CO3, 1 mole of CO2, and 1 mole of H2O.

a) To calculate the mass of residue (Na2CO3) produced, we first need to convert the given mass of NaHCO3 to moles. The molar mass of NaHCO3 is:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Total molar mass = 22.99 + 1.01 + 12.01 + (16.00 x 3) = 84.00 g/mol

Number of moles of NaHCO3 = 8.4 g / 84.00 g/mol = 0.10 mol

Since the ratio of NaHCO3 to Na2CO3 is 2:1, we can determine that 0.10 mol of NaHCO3 will produce:
0.10 mol of Na2CO3

The molar mass of Na2CO3 is:
Na = 22.99 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Total molar mass = (22.99 x 2) + 12.01 + (16.00 x 3) = 105.99 g/mol

Mass of Na2CO3 produced = 0.10 mol x 105.99 g/mol = 10.60 g

Therefore, 8.4 g of sodium hydrogen carbonate will produce 10.60 g of residue (Na2CO3).

b) To calculate the volume of carbon(IV) oxide (CO2) produced at STP (standard temperature and pressure), we can use the ideal gas law. The molar volume of a gas at STP is 22.4 L/mol.

The reaction shows that 2 moles of NaHCO3 produce 1 mole of CO2. Therefore, 0.10 mol of NaHCO3 will produce:
0.10 mol of CO2

Volume of CO2 produced at STP = 0.10 mol x 22.4 L/mol = 2.24 L

Therefore, 8.4 g of sodium hydrogen carbonate will produce 2.24 L of CO2 at STP.