Question #e7dd8

Question

Question #e7dd8

1 Answer

Impact of this question

1073 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T03:47:59

4.76 mL

Explanation:

The molar mass of $H_{2} S O_{3} = 82.01 g m o l^{- 1}$ $H_2SO_3=82.01gmol^(-1)$

Molar number of $H_{2} S O_{3} = \frac{0.0351 g}{82.01 g m o l^{- 1}} = 4.28 \times 10^{- 4} m o l$ $H_2SO_3=(0.0351 g)/(82.01gmol^(-1))=4.28xx10^(-4)mol$

Molarity of $H_{2} S O_{3} = S 1 = 4.28 \times 10^{- 4} m o l \times 1000 m l \frac{1}{250 m l} = 1.712 \times 10^{- 3} M$ $H_2SO_3=S1=4.28xx10^(-4)molxx1000ml1/(250ml)=1.712xx10^(-3)M$

Reaction:

$H_{2} S O_{3} + 2 N a O H \to N a_{2} S O_{3} + 2 H_{2} O$ $H_2SO_3+2NaOH->Na_2SO_3+2H_2O$
1mole of $H_{2} S O_{3}$ $H_2SO_3$ reacts with 2 mole NaOH.
$S u p p o s e$ $Suppose$ ,

S1=Molarity of $H_{2} S O_{3} = 1.712 \times 10^{- 3} M$ $H_2SO_3=1.712xx10^(-3)M$ ,V1=Volume of

$H_{2} S O_{3} = 250 m L$ $H_2SO_3=250 mL$

S2=Molarity of NaOH=.18M,V2=Volume of NaOH=?

$\frac{(S 1 V 1) (H_{2} S O_{3})}{(S 2 V 2) (N a O H)} = \frac{1}{2}$ $((S1V1)(H_2SO_3))/((S2V2)(NaOH))=1/2$

=)V2= $\frac{2 S 1 V 1}{S 2} = 4.76 m L$ $(2S1V1)/(S2)=4.76 mL$

Science

Math

Humanities

... and beyond

Question #e7dd8

1 Answer

Explanation:

Related questions

Impact of this question