Question

(a) How many moles of ammonium ions are in 0.552 g of ammonium carbonate? mol

(b) What is the mass, in grams, of 0.0279 mol of iron(III) phosphate? g

(c) What is the mass, in grams, of 4.98 1023 molecules of aspirin, C9H8O4? g

(d) What is the molar mass of a particular compound if 0.087 mol weighs 6.48 g? g/mol

Answer #1

Molar mass of *(NH4)2CO3* is **96.0858
g/mol**

0.552 g of ammonium carbonate = 0.552 x 2 / 96.0858 =
**0.0114889 MOles of ammonium ions presents.**

Molar mass of *FePO4* is **150.8164
g/mol**

0.0279 mol of iron(III) phosphate = 0.0279 x 150.8164 =
**4.2077 gm iron(III) phosphate presents**

4.98 1023 molecules of aspirin, C9H8O4 = 4.98 10^{23} /
6.023 x 10^{23} = 0.8268 moles presents

0.8268 moles of C9H8O4 (Molar mass of *C9H8O4* is
**180.1574 g/mol**)= 0.8268 x
**180.1574** = 1**48.9541 gm of C9H8O4
presents**

molar mass of compound = 6.48 / 0.087 = **74.48 is the the
molar mass**

(a) How many moles of ammonium ions are in 0.807 g of ammonium
carbonate?
(b) What is the mass, in grams, of 0.0445 mol of iron(III)
phosphate?
(c) What is the mass, in grams, of 2.48 1023 molecules of
aspirin, C9H8O4?
(d) What is the molar mass of a particular compound if 0.060 mol
weighs 5.34 g?

A) How many moles of oxygen atoms are in 4.48 moles of magnesium
phosphate: Mg3(PO4)2 (Mw. 262.87
g/mol)?
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hydroxide: Al(OH)3 (Mw. 78.00 g/mol)?
C) How many iron ions are in 4.5 moles of iron (III) carbonate:
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