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arsen [322]
3 years ago
11

A compound is formed when 12.2 g Mg combines completely with 5.16 g N. What is the percent composition of this compound? Please

show all work and include the percentage of both magnesium and nitrogen.
Chemistry
1 answer:
STALIN [3.7K]3 years ago
4 0
3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷<span>24,4g/mol=0,5mol-limiting reagent
</span>n(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.
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Answer:

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<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
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Explanation:

<u>Step 1: Define</u>

47.7 g Mg

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<u />47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} ) = 1.18161 × 10²⁴ particles Mg

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