Answer:
0.0707 mol of CO2.
Step-by-step explanation:
Based on the balanced chemical equation provided:
Na2CO3 + 2 HNO3 --> 2 NaNO3 + CO2 + H2O
The stoichiometric ratio between Na2CO3 and CO2 is 1:1, which means 1 mole of Na2CO3 produces 1 mole of CO2.
Given that 7.50 g of Na2CO3 reacts, we need to convert the mass of Na2CO3 to moles using its molar mass, and then use the stoichiometry to determine the moles of CO2 produced.
The molar mass of Na2CO3 is:
2(Na) + 1(C) + 3(O) = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol
Using the formula:
moles = mass / molar mass
moles of Na2CO3 = 7.50 g / 105.99 g/mol = 0.0707 mol of Na2CO3
Since the stoichiometry between Na2CO3 and CO2 is 1:1, the number of moles of CO2 produced would also be 0.0707 mol.