If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

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If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

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Answer 1 · 2016-02-18T17:23:35

Pb(NO3)2- Lead Nitrate
NaI - Sodium Iodide

$P b {(N O_{3})}_{2} + 2 N a I - \to P b I_{2} + 2 N a N O_{3}$ $Pb(NO_3)_2 + 2NaI --> PbI_2+ 2NaNO_3$

Moles of Lead(II) Nitrate = $m a s s / m o l a r m a s s$ $mass // molar mass$
= 25.0 grams//(207.20 xx 1 + 14.01 xx 2 + 16 xx 6
= 25.0 grams//331.22
= 0.075 moles

Moles of Sodium Iodide
= 15//(22.99+126.9)
= 15//149.89
= 0.100 Moles

Limiting Reagent in this case is Lead(II) Nitrate
Ratio of Moles of Lead Nitrate to Sodium Nitrate
1:2
Ratio of "moles" of lead nitrate to Sodium Nitrate
0.075 : x
$\frac{1}{2} = \frac{0.075}{x}$ $1/2 = 0.075/x$
$x = 2 \times 0.075$ $x = 2 xx 0.075$
$x = 0.150$ $x = 0.150$
Thus that means 0.150 moles of Sodium Nitrate
To calculate mass of Sodium Nitrate
Since we know that Moles = $M a s s / M o l a r M a s s$ $Mass//Molar Mass$
Thus Mass = $M o l e s \times M o l a r M a s s$ $Mol es xx Molar Mass$
Molar Mass of $N a N O_{3}$ $NaNO_3$
Na - 22.99
N - 14.01
O - 3(16) = 48
85g/mol
Thus, Mass = $0.150 \times 85$ $0.150 xx 85$
Mass = $12.75$ $12.75$
Thus, 12.75g of Sodium Nitrate can be formed

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If I start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?

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