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Lena [83]
3 years ago
5

Suppose 17. g of hydrochloric cid is mixed with 6.99 g of sodium hydroxide calculate the minimum mass of hydochloric acid taht c

ould be left over by the vhemical reaction
Chemistry
1 answer:
Eva8 [605]3 years ago
8 0

Answer: 10.62g

Explanation:

First let us generate a balanced equation for the reaction.

HCl + NaOH —> NaCl + H2O

Molar Mass of HCl= 1 + 35.5 = 36.5g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

From the question,

Mass of HCl = 17g

Mass of NaOH = 6.99g

Converting these Masses to mole, we obtain:

n = Mass / Molar Mass

n of HCl = 17/36.5 = 0.4658mol

n of NaOH = 6.99/40 = 0.1748mol

From the question,

1 mole of NaOH requires 1mole of HCl.

Therefore, 0.1748mol of NaOH will also require 0.1748mol of HCl.

But we were told that 17g( i.e 0.4658mol) of HCl were mixed.

Therefore, the unreacted amount of HCl = 0.4658 — 0.1748 = 0.291mol

Converting this to mass, we have:

Mass of HCl = n x molar Mass

Mass of HCl = 0.291 x 36.5

Mass of HCl = 10.62g

Therefore the left over Mass of HCl is 10.62g

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Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 ​ ​ 2Ag+Cu(NOX 3 ​ )X 2 ​ How many moles of \ce{Ag}
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Explanation:

<em>Given the following reaction: </em>

<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>

<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 ​ is available in excess.</em>

First, we write the balanced equation.

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We can establish the following relations.

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The moles of Ag produced from 16.0 g of Cu are:

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