Question

Calculate the moles of carbon dioxide produced at 22°C and 3 psi when 1.41 g of calcium carbonate reacts with 15.0 mL of 2M hydrochloric acid

Answer 1

Solution Ipsi = 0.068 atm 3psi = 0.204 atm Temp = 22 + 273 k = 295k RE 0.0821 atml K mo) for 1 mole of CO2 we required V(L) volume, Pr=RT VE 0.0821 x 295 0.204 v- 118.72 L Now, CaCO₃ + oHU co +40+ Call 1.41 15X2 n X1000 mol mole 0.03 n ( mo mole nxm ) 0.0141 mole mole 0.03 2 Limiting 0.0141 reagent 20.015 so limiting reagent is Calz CS Scanned with CamScanner

cor produced on the based of on the based of CaCO3 react moles of calo moles of of co v 0.0141 for 118.72 n. gases v 118.72 (2) ༡ V = 1.673 L produced Now we have have calculate at 22°C & 3psi I mole of CO₂ having 118.72 L AT SO 118.72 L co₂ I mole i L cor 11 mole 118.72 Now 1.673 L 102 1 x 1.673 116.22 mol 30, co₂ produced 0.0141 mol OR Imp to number of moles produced according how many no. of moles react of cacoz so so moles of Calo, 0.0141 moles of Wz produced also 0.0141 And CS Scanned with CamScanner

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