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2 years ago

Calculate the theoretical yield if 100.0 g p4o10 react with 200.0 g h2o

1 answer:

3 0

The reaction of Phosphorous Pentaoxide with water yield Phosphoric Acid as shown below,

P₄O₁₀ + 6 H₂O → 4 H₃PO₄

According to balance equation,

283.88 g (1 mole) P₄O₁₀ requires = 108 g (6 mole) of H₂O
So,
100 g P₄O₁₀ will require = X g of H₂O

Solving for X,

X = (100 g × 108 g) ÷ 283.88 g

X = 38.04 g of H₂O

So, 100 g P₄O₁₀ requires 38.04 g of H₂O, while we are provided with 200 g of H₂O which means that water is in excess and P₄O₁₀ is limiting reagent. Therefore, P₄O₁₀ will control the yield of H₃PO₄. So,
As,
283.88 g (1 mole) P₄O₁₀ produced = 391.96 g (4 mole) of H₃PO₄
So,
100 g P₄O₁₀ will produce = X g of H₃PO₄

Solving for X,
X = (100 g × 391.96 g) ÷ 283.88 g

X = 138.07 g of H₃PO₄

Result:
Theoretical Yield of this reaction is 138.07 g.

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A pound is approximately 0.45 kilogram. A person weighs 87 kilograms. What is the person’s weight, in pounds, when expressed to

Sauron [17]

Answer:

87 kilograms =191.8 pounds

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2 years ago

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Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na

Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

$\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}$

valence according to the amount of salt anion

Input the value :

$\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3$

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2 years ago

A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

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Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2

maxonik [38]

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$ .... $(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$ .... $(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

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2 years ago

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How many grams of (nh4)3po4 are needed to make 0.250 l of 0.150 m (nh4)3po4? hints how many grams of (nh4)3po4 are needed to mak

NeX [460]

<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.

Moles </span>(nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4

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2 years ago