Question #9e3a5

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Answer 1 · 2017-11-22T02:31:51

Working from the following metathesis reaction ...

$2HNO_3(aq) + Ca(OH)_2(aq)$ => $Ca(NO_3)_2 + 2H_2O(l)$

The reaction ratio to neutralization is 2 moles $HNO_3$ to 1 mole $Ca(OH)_2$

From definition of
$"Molarity"(M) = ("moles"."solute")/("Liters"."solution")$

=> $"moles"."solute"= "Molarity(M)"xx"Volume(L)"$

Therefore, one can use the relationship

1 mole $HNO_3$ = 2 moles $Ca(OH)_2$

(Molarity x Volume) $HNO_3$ = 2(Molarity x Volume) $Ca(OH)_2$

$(0.177M)HNO_3(aq) xx(28.2ml)HNO_3 "solution")$ = $2[Ca(OH)_2(aq)] xx(17.5ml)Ca(OH)_2 "solution")$

Simplifying notations for illustration

(0.177M)(28.2ml) = $2[Ca(OH)_2(aq)]$ (17.5ml)

$[Ca(OH)_2]$ = $((2(0.177M)(28.2ml))/(17.5ml))$

= $0.570M Ca(OH)_2(aq) "solution"$

Note: In this problem case, the volume values may remain in milliliter dimensions as both sides of the (MV)acid = (MV)base are in ml's.