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Question

erences Edit View d Help stchas.instructure.com ssions D 1 Question 1 es 2 pts Time Running: Hide Attempt duer Apr 7 at 12p 43 Minutes, 42 Secons le A hydrate of ZnSOs is placed in a test tube with a total mass of 54 23 & Atter heating, the mass s 38.69 g. If the test tube weighs 1883 & what is the formuls of the hydrate es labus izzes odules ZnSO4 + 7H20 Zn504.SH20 Zn504 6H20 Chat Google Drive Office 365 2 pts D : Question 2 A 20.26-8 sample of manganese metal is heated with onygen to produce 26.16 g of a manganese oxide compound. What is the empirical formula of the compound? MnOz Mio

Explanation / Answer

1)

mass of hydrate before heating = 54.23 g - mass of tube

= 54.23 g - 18.83 g

= 35.40 g

mass of anyhydrous = 38.69 g - mass of tube

= 38.69 g - 18.83 g

= 19.86 g

mass of H2O = mass of hydrated salt - mass of anhydrous salt

mass of H2O = 35.40 g - 19.86 g

mass of H2O = 15.54 g

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = 15.54 g

we have below equation to be used:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(15.54 g)/(18.016 g/mol)

= 0.8626 mol

Molar mass of ZnSO4 = 1*MM(Zn) + 1*MM(S) + 4*MM(O)

= 1*65.38 + 1*32.07 + 4*16.0

= 161.45 g/mol

mass of ZnSO4 = 19.86 g

we have below equation to be used:

number of mol of ZnSO4,

n = mass of ZnSO4/molar mass of ZnSO4

=(19.86 g)/(161.45 g/mol)

= 0.123 mol

we have below equation to be used:

X = mol (H2O)/mol (ZnSO4)

X = 0.8626 / 0.123

X = 7

Answer: ZnSO4.7H2O (Option b)

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