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erma4kov [3.2K]
3 years ago
8

Calculate the volume in milliliters of a 1.3M zinc nitrate solution that contains 100.g of zinc nitrate ZnNO32. Be sure your ans

wer has the correct number of significant digits.
Chemistry
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

The volume is 406 mL

Explanation:

Step 1: Data given

Molarity of a zinc nitrate solution = 1.3 M

Mass of zinc nitrate = 100 grams

Molar mass of Zn(NO3)2 = 189.36 g/mol

Step 2: Calculate moles Zn(NO3)2

Moles Zn(NO3)2 = mass Zn(NO3)2 / molar mass Zn(NO3)2

Moles Zn(NO3)2 = 100.0 grams / 189.36 g/mol

Moles Zn(NO3)2 = 0.5281 moles

Step 3: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.5281 moles / 1.3 M

Volume = 0.406 L = 406 mL

The volume is 406 mL

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20.00 g of aluminum (Al) reacts with 78.78 grams of molecular chlorine (Cl2), all of each reaction is completely consumed and as
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The reaction forms 98.76 g AlCl_3.  

We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1. Gather all the information</em> in one place with molar masses above the formulas and everything else below them.  

M_r: ___26.98 _70.91 __133.34

________2Al + 3Cl_2 → 2AlCl_3

Mass/g: 20.00 _78.78

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

Moles of Al = 20.00 g Al × (1 mol Al /26.98 g Al) = 0.741 29 mol Al

Moles of Cl_2 = 78.78 g Cl_2 × (1 mol Cl_2 /70.91 g Cl_2) = 1.11 10 mol Cl_2

Step 3. Identify the <em>limiting reactant</em>  

Calculate the moles of AlCl_3 we can obtain from each reactant.  

<em>From Al</em>: Moles of AlCl_3 = 0.741 29 mol Al × (2 mol AlCl_3/2 mol Al) = 0.741 29 mol AlCl_3

<em>From Cl_2</em>: Moles of AlCl_3 = 1.11 10 mol Cl_2 × (2 mol AlCl_3/3 mol Cl_2) = 0.740 66 mol AlCl_3

<em>Cl_2 is the limiting reactant</em> because it gives the smaller amount of AlCl_3.

<em>Step 4</em>. Calculate the <em>mass of AlCl_3</em>.

Mass = 0.740 66 mol AlCl_3 × 133.34 g/1 mol AlCl_3 = 98.76 g AlCl_3

The reaction produces 98.76 g AlCl_3.

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