M State . CHEM 1100 . ng.com/ibiscms/mod/ibis/view.php?id-4517923 Jump to... Whe

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M State . CHEM 1100 . ng.com/ibiscms/mod/ibis/view.php?id-4517923 Jump to... When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate are combined with 13.0 g of hydrochloric acid? Number 22.2 & CaC, Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete? Number g of Previous Give Up & View Solution Cheak Answer Next Ext earch

Explanation / Answer

Nnumber of moles of calcium carbonate = 29.0 g / 100.0869 g/mol = 0.290 mol

number of moles of HCl = 13.0 g / 36.46 g/mol = 0.357 mol

from the balanced equation we can say that

1 mole of CaCO3 requires 2 mole of HCl so

0.290 mole of CaCO3 will require

= 0.290 mole of CaCO3 *(2 mole of HCl / 1 mole of CaCO3)

= 0.580 mole of HCl

But we have only 0.357 mol of HCl so HCl is the limiting reactant

from the balanced equation we can say that

2 mole of HCl produces 1 mole of CaCl2 so

0.357 mole of HCl will produce 0.179 mole of CaCl2

mass of 1 mole of CaCl2 = 110.98 g

so the mass of 0.179 mole of CaCl2 = 19.9 g

Therefore, the mass of CaCl2 produced would be 19.9 g

excess reactant is CaCO3

from the balanced equation we can say that

2 mole of HCl requires 1 mole of CaCO3 so

0.357 mole of HCl will require 0.179 mole of CaCO3

number of moles of excess reactant remain = 0.290 - 0.179 = 0.111 mole

Therfore, the mass of excess reactant remain after reaction is 0.111 * 100.0869 = 11.1 g

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