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What is the mass of 1.45 moles of silver sulfate?

User Coykto
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2 Answers

4 votes
The answer is 449.5g
User Sdfx
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7 votes

Answer:

449.5 g

Step-by-step explanation:

Silver sulfate- Ag2SO4

M(Ag)=107 g/mol => M(Ag2)=214 g/mol

M(S)=32 g/mol

M(O)=16 g/mol => M(O4)=64 g/mol

M(Ag2SO4)=310 g/mol

n=1.45 mol

m(Ag2SO4)=M(Ag2SO4)*n=310 g/mol *1.45 mol= 449.5 g

User Nikita Volkov
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