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MA_775_DIABLO [31]
3 years ago
11

How much 6.0M HNO(small 3) is needed to neutralize 39mL of 2 M KOH

Chemistry
1 answer:
Digiron [165]3 years ago
8 0
Equation of reaction 
<span>KOH+HNO3--->KNO3+H2O </span>
<span>CA=6.0M, CB=2.0M, VA=?, VB=39ml, na=1, nb=1 </span>
<span>CAVA/CBVB=na/nb </span>
<span>6*VA/2*39=1/1 </span>
<span>6VA/78=1 </span>
<span>6VA=78 </span>
<span>VA=78/6. VA=13ml. </span>
<span>VA=13ml.</span><span>
Hope this helped!!!!! </span>Don't forget Brainliest!!!!! You can always PM me if you don't want to waste any points!!!!!
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Explanation: For the calculation of moles, we use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}     ....(1)

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Putting values in equation 1, we get:

Moles=\frac{92g}{28g/mol}=3.285moles

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Given mass = 112 grams

Molar mass = 116g/mol

Putting values in equation 1, we get:

Moles=\frac{112g}{116g/mol}=0.965moles

The reaction follows:

2Cr_2O_3(s)+3Si(s)\rightarrow 4Cr(l)+3SiO_2(s)

By Stoichiometry,

2 moles of Cr_2O_3 reacts with 3 moles of silicon

So, 0.965 moles of Cr_2O_3 reacts with = \frac{3}{2}\times 0.965 = 1.4475 moles of Silicon.

As, the moles of silicon is more than the required amount and is present in excess.

So, the excess reagent for the reaction is Silicon.

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