Convert 345 NH3 grams to moles

asked Jan 1, 2018 2.5k views

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1 vote

m(NH₃)=345 g

M(NH₃)=A(N)+3A(H)
M(NH₃)=14.0+3·1.0=17.0g/mol

n(NH₃)=m(NH₃)/M(NH₃)

n(NH₃)=345g/17.0g/mol=20.3 mol

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2 votes

345 NH3 in moles is: 20.2577490294015 moles.

Derwyn answered Jan 8, 2018

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