85 grams of CuCl2 are reacted with 186 grams of NaNO3. How many moles of NaCl are produced? Answer to two decimal places. (Hint: find the limiting reactant first!) CuCl2 + 2 NaNO3 → Cu(NO3)2 + 2 NaCl CuCl2 = 134.45 g/mol %3D NaNO3 = 84.99 g/mol
Limiting reagent :
The reactant limits the amount of product that is formed by the chemical reaction. This limiting reagent completely consumed during the reaction and then it will stop the reaction and no more product will form.
CuCl2 + 2 NaNO3Cu(NO3)2+ 2 NaCl
Given reaction is balanced. In this case, we need to find out the limiting reactant first and for that, we need to find how many moles of CuCl2 and NaNO3 are given and then what will be its stoichiometric ratio. For every 2 moles of NaNO3, we required 1 mole of CuCl2.This is what the equation tells us.
Now, we need to find out the molecular weight of CuCl2,
Molecular weight of CuCl2= {(molecular weight of Cu)+ (2molecular weight of chlorine)}
= {(63.546) +( 235.45)}
= 134.45 g
1 mole of NaNO3 ={(molecular weight of Na)+ (Molecular weight of nitrogen)+(3molecular weight of oxygen)}
= 85g
1 mole of NaCl ={(molecular weight of Na)+ (molecular weight of chlorine)}
= 58.44 g
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