Answer:
a) 0.925 mol Na2CO3 can be theoretically produced
b) 0.075 moles of the excess starting material remains
Step-by-step explanation:
balaced chemical:
2 NaOH(s) + CO2(g) ↔ Na2CO3(s) + H2O(aq)
1.85n 1.00n Xn
moles theor. Na2CO3:
⇒ nNaCO3 = 1.85nNaOH * ( nNa2CO3 / 2nNaOH)
⇒ nNa2CO3 = 0.925nNa2CO3
moles of the excess:
⇒moles CO2 react = 1.85nNaOH * nCO2 / 2nNaOH = 0.925n CO2
⇒moles CO2 excess = 1.00n - 0.925n = 0.075n excess CO2