Question

How many grams of ca(no3)2 can be produced by reacting excess hno3 with 6.33 g of ca(oh)2?

Answer 1

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

$Moles\ Ca(OH)2 = \frac{Mass}{Molar\ Mass} = \frac{6.33 g}{74 g/mol} =0.0855$

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

$Mass\ Ca(NO3)2 \ produced = moles*molar\ mass \\= 0.0855\ moles*164\ g/mol = 14.02\ g$

Answer 2

14.022 grams of Ca(NO₃)₂.

Further explanation

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

• Relative atomic mass: Ca = 40, N = 14, O = 16, and H = 1.
• Relative molecular mass (Mr) of Ca(OH)₂ = 40 + 2(16 + 1) = 74.
• Relative molecular mass (Mr) of Ca(NO₃)₂ = 40 + 2[14 + 3(16)] = 164.

Let us convert mass to mol of 6.33 g of Ca(OH)₂.

$\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{6.33}{74} = 0.0855 \ moles \ }$

Step-2

Balanced reaction:

$\boxed{ \ Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O\ }$

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.

$\boxed{ \ \frac{n(Ca(NO_3)_2)}{n(Ca(OH)_2)} = \frac{1}{1} \ } \rightarrow \boxed{ \ n(Ca(NO_3)_2) = n(Ca(OH)_2) \ }$

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.

$\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ mass = n \times Mr \ }$

$\boxed{ \ n = 0.0855 \ moles \times 164 \ \frac{g}{mol} \ }$

Thus, we can produce 14.022 grams of Ca(NO₃)₂.

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