2.what volumen of 0.612 M ammonium sulfate contains 6.70g of ammonium ion? 3.how

Question

Explanation / Answer

2.

Ammonium sulphate is a salt and hence dissociates fully into ammonium and sulphate ions

(NH4)2SO4 ---------> NH4+ + SO42-

1 MOLE 1 mole

So, 0.612 moles of (NH4)2SO4 ---------------> 0.612 moles of NH4+ ions too

Molarity = moles

Volume in litrres

and moles = grams/molar mass

So,

0.612 moles/litre = 6.70 g / (18 g/mole)

x litres

0.612 moles/litre = 0.3722 moles

x litres

x = 0.6082 litres

or x = 608.2 ml of NH4+ ion

3.

Ba(OH)2 + 2HCl --------->BaCl2 + 2H2O

1 mole 2 moles

M1 V1 = M2 V2

n1 n2

0.0365 * 0.231 = 0.283 * x L

1 2

0.0084315 = 0.283 * x

2

x = 0.05958 L

x = 59.6 ml [answer] - 3 significant figures as in 0.283 ml [3 significant figures]

4.

CaCl2 + 2AgNO3 ------> 2AgCl + Ca(NO3)2

1 mole 2 moles 2 moles

Moles of CaCl2 = 0.150 moles/litre * 0.030 litre

= 0.0045 moles

Moles of AgNO3 = 0.100 moles/litre * 0.0135 litre

= 0.00135 moles

0.00135 moles AgNO3 * 1 mole CaCl2 = 0.000675 moles CaCl2

1 2 moles AgNO3

While we have a total of 0.45 moles of CaCl2, only 0.000675 moles are used [consumed] for the 0.00135 moles of AgNO3 as per the mole:mole 1:2 ratio

So CaCl2 is in excess and hence AgNO3 is the limiting reagent and moles of the product AgCl would be formed based on the moles of limiting reagent AgNO3

since its 2:2 ratio [1:1 ratio in simple] for AgNO3 and AgCl

number of moles of AgCl formed woudl be same as that of AgNO3 = 0.00135

Grams of AgCl = moles * molar mass

= 0.00135 * 143.32

= 0.1935 grams

30.0 and 0.150 are 3 significant figures

so final answer, grams of ASgCl formed = 0.194 g

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