Useful information: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00g/mol, Ca = 40.08

Question

Useful information: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00g/mol, Ca = 40.08 g/mol, Cl = 35.45 g/mol

What volume, in mL, of 0.7376 molar hydrochloricacid is required to react with 1.1835 g of calcium carbonate?Assume a 100% yield and report your answer to the correct number ofsignificant digits. DO NOT enter units as part ofyour answer.

Useful information: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00g/mol, Ca = 40.08 g/mol, Cl = 35.45 g/mol

2 HCl (aq) + CaCO3 (aq)---> CaCl2 (aq) + CO2(g) + H2O (l)

Explanation / Answer

2 HCl (aq) + CaCO3(aq) ---> CaCl2 (aq) +CO2 (g) + H2O (l) No . of moles of CaCO3 is , n = mass / Molarmass Molar mass of CaCO3 is = 40 + 12 +3 * 16 =100 g So , n = 1.1835 / 100           =0.011835 moles 1 mole of CaCO3 requires 2 moles of HCl 0.011835 moles of CaCO3 requires 2 * 0.011835(=0.02367) moles of HCl No . of moles = Molarity * Volume in L 0.02367   = 0.7376 * V V = 0.032 L     = 32 mL

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