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Question 1 (5 points)
When 45.8g sodium chloride reacts with 75.0g Calcium nitrate, Calcium chloride and sodium nitrate
forms. How many grams of calcium chloride and sodium nitrate are produced?
NaNO3 + 1
CaCl2
2. NaCl + 1 Ca(NO3)2 - 2
Mass of Sodium Nitrate Produced
(round to one decimal place)
(round to one decimal place)
Mass of Calcium Cloride Produced
Limiting Reagent for the reaction

Sagot :

Oseniidn

The mass of sodium nitrate and calcium chloride produced would be 66.3 g and 43.3 g respectively.

Stoichiometric reaction

The equation of the reaction between sodium chloride and calcium nitrate is given by the following equation:

2NaCl + Ca(NO3)2 ---> CaCl2 + 2NaNO3

Mole ratio of NaCl and Ca(NO3)2 = 2:1

Mole of 45.8g NaCl = 45.8/58.44

                                 = 0.78 moles

Mole of 75.0 g Ca(NO3)2 = 75/164.1

                                         = 0.46 moles

Equivalent mole of Ca(NO3)2 = 0.78/2 = 0.39

Ca(NO3)2 is in excess. Thus NaCl is the limiting reactant.

Mole ratio of NaCl and CaCl2 = 2;1

Equivalent mole of CaCl2 = 0.78/2 = 0.39

Mass of 0.39 CaCl2 = 0.39 x 110.98

                                 = 43.3 g

Mole ratio of NaCl and NaNO3 = 1:1

Equivalent mole of NaNO3 = 0.78/1 = 0.78

Mass of 0.78 mole NaNO3 = 0.78 x 84.99

                                           = 66.3 g

More on stoichiometric equations can be found here: https://brainly.com/question/12733510

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