# Using the following information to help you answer parts (a) - (c): HNO3 + NaOH ---> NaNO3 + H2O ?

## RAM: H = 1 NA = 23 0 = 16 N = 14 A) if you have 0.55 moles of HNO3 how many oxygen atoms would you have? B) If 0.365g of NaOH is reacted with an excess of HNO3 what is the maximum mass of NaNO3 which could be recovered? C) 12.05cm3 of a 0.2065 M aqueous NaOH solution is titrated to the end-point by 25.05 cm3 of HNO3 solution. What is the concentration of the HNO3?

Apr 14, 2018

$\text{ The given equation is } H N {O}_{3} + N a O H \to N a N {O}_{3} + {H}_{2} O$

We, can conclude that, $1 \text{ mole of } H N {O}_{3}$ reacts with $1 \text{ mole of } N a O H$ to produce $1 \text{ mole of } N a N {O}_{3}$ and $1 \text{ mole of } {H}_{2} O .$

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A) If I have $0.55 \text{ moles of } H N {O}_{3}$ how many oxygen atoms would I have?

$1 \text{ mole of } H N {O}_{3}$ has $6.022 \times {10}^{23} \text{ molecules of } H N {O}_{3}$

$0.55 \text{ moles of } H N {O}_{3}$ would have $6.022 \times 0.55 \times {10}^{23} \text{ molecules of } H N {O}_{3}$

$\implies 3.3121 \text{ molecules of } H N {O}_{3}$

Each molecule of $H N {O}_{3}$ contains 3 Oxygen atom. Hence the total number of oxygen atoms would be, $3.3121 \times 3$

=> color(red)( 9.93 atoms of oxygen are present in $0.55 \text{ moles of } H N {O}_{3}$

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B) If I react $0.365 \text{g of } N a O H$ with an excess of $H N {O}_{3}$ what is the maximum mass of $N a N {O}_{3}$ that would be recovered?

number of mole of $N a O H$ = $\text{given mass"/"molar mass}$

$\implies \frac{0.365}{40} = 0.0091 \text{ moles}$

As $1 \text{ mole of } N a O H$ to produces $1 \text{ mole of } N a N {O}_{3}$,
$0.0091 \text{ moles of } N a O H$ to produces $0.0091 \text{ moles of } N a N {O}_{3}$

$\implies 0.0091 \times 85$

=> color(magenta)(0.7735g

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C) $12.05 c {m}^{3}$ of a $0.2065 M$ aqueous $N a O H$ solution is titrated to the end-point by $25.05 c {m}^{3}$ of $H N {O}_{3}$ solution. What is the concentration of the $H N {O}_{3}$?

Let,

${M}_{1} = 0.2065 M$ color(white)(rrrrwwwrrrr ${V}_{1} = 12.05 c {m}^{3}$

M_2= ? M color(white)(rrrwwwwwwrrrrr ${V}_{2} = 25.05 c {m}^{3}$

By the formula, ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

$\implies \left(0.2065 M\right) \left(12.05 c {m}^{3}\right) = {M}_{2} \left(25.05 c {m}^{3}\right)$

=> color(blue)(M_2=0.099M