Question

011214
*49. Find the amount of the indicated element (in moles) in
(a) 8.75 g of B203.
(b) 167.2 mg of Na2B40, 10H,O.
(c) 4.96 g of Mn304.
(d) 333 mg of CaC204.​

Answer 1

Answer:

a) 8.75 g of B2O3

molar mass= 11×2+16×3 g

= 22+48 g

= 70 g

given mass= 8.75g

no. of moles= given mass/ molar mass

= 8.75/70

= 0.125 moles

b)167.2 mg of Na2B4O,10H,O

molar mass of Na2B4O= 23×2+11×4+16 g

= 46+44+16 g

= 106 g

molar mass of 10H= 10×1 g = 10 g

molar mass of O= 1×16 = 16 g

given mass= 167.2 g

no. of moles= given mass/molar mass

= 167.2/106 = 1.577 moles

= 167.2/10 = 16.72 moles

= 167.2/16 = 10.45 moles

c) 4.96 g of Mn3O4

molar mass of Mn3O4= 55×3+16×4 g

= 165+64 g

= 229 g

given mass= 4.96 g

no. of moles= given mass/molar mass

= 4.96/229

= 0.02 moles

Answer 2

8.75g of B2O3 8.75g 340