Chemistry, asked by prashantjangade779, 1 day ago

415 mL of 0.275 M copper(II) nitrate solution is reacted to 25.8 g of sodium bicarbonate (NaHCO3) to form copper(II) carbonate, sodium nitrate, water, and carbon dioxide. If reaction took place with 75% yield, how many grams of copper (II) cabonate forms?

Answers

Answered by bhatmubashir090
0

Explanation:

Correct option is

D

57 g

According to law of conservation of mass, total mass of reactants is equal to total mass of products.  Here baking soda mixture on heating gives solid residue and carbon dioxide.

MBaking soda mixture=Msolid residue+MCO2

or it can be written as:

Msolid residue=MBaking soda mixture−MCO2  (Eq1) 

According to the question

MBaking soda mixture=100g

MCO2=43g

Substituting the value in equation 1, we get

The mass of solid residue is 100g−43g=57g

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