8.4g of sodium hydrogen carbonate is completely decomposed by heat.Calculate:
a)the mass of residue produced
b)Volume of carbon(IV)oxide produced at stp
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.