When 34.2 grams of strontium reacts with excess molecular oxygen, how many grams of the product...
When 34.2 grams of strontium reacts with excess molecular oxygen, how many grams of the product will form?
Homework Answers
Molar mass of Sr = 87.62 g/mol
mass of Sr = 34.2 g
mol of Sr = (mass)/(molar mass)
= 34.2/87.62
= 0.3903 mol
Balanced chemical equation is:
2 Sr + O2 —> 2 SrO
According to balanced equation
mol of SrO formed = moles of Sr
= 0.3903 mol
Molar mass of SrO,
MM = 1*MM(Sr) + 1*MM(O)
= 1*87.62 + 1*16.0
= 103.62 g/mol
mass of SrO = number of mol * molar mass
= 0.3903*1.036*10^2
= 40.45 g
Answer: 40.5 g
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