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3 years ago

How many grams of H,O are needed to produce 13 g of NaOH? 2 Na,O2 + 2 H20-4 NaOH + O2

Chemistry

1 answer:

fgiga [73]3 years ago

3 0

Answer:

This is a typical stoichiometry question.To answer this question you want to get a relationship between

O and NaOH.

So you can get a relationship between the moles of

and moles of NaOH by the concept of stoichiometry.

O +

O ----------------> 2 NaOH.

According to above balanced equation we can have the stoichiometry relationship between

O and NaOH. as 1:2

It means 1 moles of

O is required to react with one mol of

O to produce 2 moles of NaOH.

in terms of mass 1 mole of

O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.

62 g of

O produces 80 g of NaOH.

1g of NaOH is produced from 62/80 g of

1.6 x

g of NaOH will require 62 x 1.6 x

g / 80 of

124g of

Explanation:

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3 years ago

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

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Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles$

$\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles$

$Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4$

According to stoichiometry :

1 mole of $Ca(NO_3)_2$ require = 1 mole of $Li_2SO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ will require= $\frac{1}{1}\times 0.061=0.061moles$ of $Li_2SO_4$

Thus $Ca(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Li_2SO_4$ is the excess reagent.

As 1 mole of $Ca(NO_3)_2$ give = 1 mole of $CaSO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ give = $\frac{1}{1}\times 0.061=0.061moles$ of $CaSO_4$

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3 years ago

Gifblaar is a small South African shrub and one of the most poisonous plants known because it contains fluoroacetic acid (FCH2CO

PSYCHO15rus [73]

[H_{3}O^{+}] = 0.00770 M

The equilibrium equation representing the dissociation of $FCH_{2}COOH$

$FCH_{2}COOH(aq) + H_{2}O (l) FCH_{2}COO^{-}(aq)+ H_{3}O^{+}(aq)$

Given [H_{3}O^{+}] = 0.00770 M

Let the initial concentration of acid be x and change y

So y = $[H_{3}O^{+}]$ = $[FCH_{2}COO^{-}]$ = 0.00770 M

$pK_{a} = 2.59K_{a} = 10^{-2.59} = 0.00257 M$

$K_{a} = \frac{(0.00770 M)(0.00770 M)}{x - 0.00770}$

$0.00257 = \frac{0.00005929}{x - 0.00770}$

0.00257 x - 0.00001979 = 0.00005929

x = 0.031 M

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3 years ago

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