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Ad libitum [116K]
2 years ago
6

I AMMMM GIVING 45 POINTSSSS PLSSS HELPPPPPPPP

Chemistry
1 answer:
Westkost [7]2 years ago
8 0

Answer:

mNaNO3 =765g

Explanation:

First, we write the balanced chemical equation representing the chemical reaction that happened between aluminum nitrate and sodium chloride.

Balanced chemical equation:

AL(NO3)3+3NaCl ⟶ 3 NaNO3+AlCl3

According to the equation, each mole of aluminum nitrate requires three moles of sodium chloride. Thus, the required number of moles of sodium chloride is

4 Mol ⋅ 3 = 12mol

Based on the data provided in the table, there were 9 moles of sodium chloride used in the reaction, which was not enough for the entirety of aluminum nitrate to react. So, sodium chloride must have been the limiting reactant.

Therefore, we use the number of moles (n) of sodium chloride to calculate the number of moles of sodium nitrate, which has a 1:1 ratio with sodium chloride.

Number of moles sodium nitrate:

nNaNO3=nNaCl

nNaNO3 = 9 mol

We can also calculate the mass (m) of sodium nitrate that was produced by multiplying its number of moles by its molar mass (MM), 85.00g/mol.

Mass of sodium nitrate produced:

mNaNO3 = nNaNO3 ⋅ MMNaNO3

mNaNO3 = 9 mol ⋅ 85.00 g/mol

mNaNO3 =765g

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sp2606 [1]

Answer:

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Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

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Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
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