<span>Fe(OH)3(S) +3HNO3(aq)----->Fe(NO3)3(aq) + 3H20(aq)
M(Fe(OH)3)=56+48+3=107; M(HNO3)= 48+14+1=63
n(Fe(OH)3)=5.4/107=0.05; n(HNO3)=2.6/63=0.04
n(Fe(OH)3):n(HNO3)=1:3, which means that the HNO3 should be three times (molar) than the Fe(OH)3, but you can see that it is, actually, even less than the Fe(OH)3, meaning that HNO3 is the limiting reagent and the amount of Fe(OH)3 which is going to react with HNO3 is 0.04/3=0.013 i.e. 0.05-0.013=0.037 mol Fe(OH)3 is left after the completion.
Just in case you can convert it into mass, but I suppose this is enough.</span>
Sorry idk. i was trying to figure it out but i cant
Do you mind submitting a picture of the table of properties you are using!
Glad to help!
The number of moles of C present in C₅H₁₂ that contains 22.5 g of H is 9.375 moles
<h3>How to determine the mass of C₅H₁₂ that contains 22.5 g of H</h3>
1 mole of C₅H₁₂ = (12×5) + (1×12) = 72 g
Mass of H in 1 mole of C₅H₁₂ = 12 × 1 = 12 g
Thus,
12 g of H is present in 72 g of C₅H₁₂
Therefore,
22.5 g of H will be present in = (22.5 × 72) / 12 = 135 g of C₅H₁₂
<h3>How to determine the mole of C present in 135 g of C₅H₁₂</h3>
72 g of C₅H₁₂ contains 5 moles of C
Therefore,
135 g of C₅H₁₂ will contain = (135 × 5) / 72 = 9.375 moles of C
Thus, 9.375 moles of C is present in C₅H₁₂ that contains 22.5 g of H
Learn more about mole:
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Answer:
Your answer should be 15.68 grams.
Explanation:
Seeing as 1 mole has a mass of 56 g, 56*0.28 would get you 15.68 g.
