Answer:
Amount of Ca(NO3)2 produced = 14.02 g
Explanation:
The given reaction can be depicted as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2
Now, Mass of Ca(OH)2 = 6.33 g
Molar mass of Ca(OH)2 = 74 g/mol
[tex]Moles\ Ca(OH)2 = \frac{Mass}{Molar\ Mass} = \frac{6.33 g}{74 g/mol} =0.0855[/tex]
Based on the reaction stoichiometry:
1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2
Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles
Molar mass of Ca(NO3)2 = 164 g/mol
[tex]Mass\ Ca(NO3)2 \ produced = moles*molar\ mass \\= 0.0855\ moles*164\ g/mol = 14.02\ g[/tex]
14.022 grams of Ca(NO₃)₂.
Given:
The reaction between excess HNO₃ with 6.33 g of Ca(OH)₂ produces Ca(NO₃)₂.
Question:
How many grams of Ca(NO₃)₂ can be produced?
The Process:
Step-1
Let us convert mass to mol of 6.33 g of Ca(OH)₂.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{6.33}{74} = 0.0855 \ moles \ }[/tex]
Step-2
Balanced reaction:
[tex]\boxed{ \ Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O\ }[/tex]
Due to excess HNO₃, then Ca(OH)₂ become limiting reagents. The number of moles of Ca(OH)₂ determines the number of moles of Ca(NO₃)₂ as a result.
According to chemical equation above, proportion between Ca(OH)₂ and Ca(NO₃)₂ is 1 to 1. Therefore, we can count the number of moles of Ca(NO₃)₂.
[tex]\boxed{ \ \frac{n(Ca(NO_3)_2)}{n(Ca(OH)_2)} = \frac{1}{1} \ } \rightarrow \boxed{ \ n(Ca(NO_3)_2) = n(Ca(OH)_2) \ }[/tex]
Hence we get 0.0855 moles of Ca(NO₃)₂.
Step-3
Let us convert mol to mass of 0.0855 moles of Ca(NO₃)₂.
[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ mass = n \times Mr \ }[/tex]
[tex]\boxed{ \ n = 0.0855 \ moles \times 164 \ \frac{g}{mol} \ }[/tex]
Thus, we can produce 14.022 grams of Ca(NO₃)₂.
