Problem 9.15 Part A Calcium carbonate reacts with HCl according to the following

Question

Problem 9.15 Part A Calcium carbonate reacts with HCl according to the following equation: 2HCl(aq) +CaCOs(s)-CaCl2(aq)+H20(1) + CO2(9) How many moles of HCl are in 61 mL of 0.13 M HCI? Express your answer using two significant figures. n= 7.9x10-3 mol Submit My Answers Give Up All attempts used; correct answer displayed Part B What mass of calcium carbonate is needed for complete reaction with the HCl in (a)? Express your answer to two significant figures and include the appropriate units m 0.49 g Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

PART B:

from the eqaution,

2 moles of HCl combines with 1 mole CaCO3

then, 7.9*10-3 moles of HCl combines with (1 * 7.9*10-3) / 2 = 3.95 * 10-3 moles of CaCO3

mass of CaCO3needed = 3.95 * 10-3 moles * 100 g/mol = 0.395 grams

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