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Given 5.0 grams of lead (II) nitrate and 3.0

grams of sodium iodide, how many grams of sodium nitrate can be formed?

User Blabdouze
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1 Answer

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Answer:

1.70 g of NaNO

Step-by-step explanation:

The balance chemical equation for given double displacement reaction is,

Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: Calculate moles of each reactant:

Moles = Mass / M.Mass

For Pb(NO)₂:

Moles = 5.0 g / 331.21 g/mol

Moles = 0.0150 moles

For NaI:

Moles = 3.0 g / 149.89 g/mol

Moles = 0.020 moles

Step 2: Calculate Limiting reagent as;

According to equation,

1 mole of Pb(NO₃)₂ reacts with = 2 moles of NaI

So,

0.0150 moles of Pb(NO₃)₂ will react with = X moles of NaI

Solving for X,

X = 2 mol × 0.0150 mol / 1 mol

X = 0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: Find out moles of NaNO₃ formed:

According to equation,

2 mole of NaI produced = 2 moles of NaNO₃

So,

0.020 moles of NaI will produce = X moles of NaNO₃

Solving for X,

X = 2 mol × 0.020 mol / 2 mol

X = 0.020 moles of NaNO₃

Step 4: Calculate Mass of NaNO₃ produced;

As, Moles = Mass / M.Mass

Or,

Mass = Moles × M.Mass

Putting values,

Mass = 0.020 mol × 84.99 g/mol

Mass = 1.70 g of NaNO

User Lanayx
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