In the reaction $P_{4} + 5 O_{2} \to P O_{4} O_{10}$ $P_4 + 5O_2 -> PO_4O_10$ , 7.75 g $P_{4}$ $P_4$ reacts. How do you calculate the mass of $P_{4} O_{10}$ $P_4O_10$ produced?

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In the reaction $P_{4} + 5 O_{2} \to P O_{4} O_{10}$ $P_4 + 5O_2 -> PO_4O_10$ , 7.75 g $P_{4}$ $P_4$ reacts. How do you calculate the mass of $P_{4} O_{10}$ $P_4O_10$ produced?

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Answer 1 · 2017-03-03T21:19:22

We follow the stoichiometric equation..........and get approx. $4 \cdot g$ $4*g$ .

Explanation:

$P_{4} (s) + 5 O_{2} (g) \to P_{4} O_{10} (s)$ $P_4(s) + 5O_2(g) rarr P_4O_10(s)$

$P^{0}$ $P^(0)$ is oxidized up to $P^{V}$ $P^(V)$ .

$Moles of white phosphorus:$ $"Moles of white phosphorus:"$

$=$ $=$ $\frac{7.75 \cdot g}{4 \times 30.9737 \cdot g \cdot m o l^{- 1}} = 6.26 \times 10^{- 2} \cdot m o l .$ $(7.75*g)/(4xx30.9737*g*mol^-1)=6.26xx10^-2*mol.$

And thus, given excess dioxygen (reasonable!), we can calculate the mass of oxide as:

$\frac{1}{4} \times 6.26 \times 10^{- 2} \cdot m o l \times 283.89 \cdot g \cdot m o l^{- 1} = ? ? g$ $1/4xx6.26xx10^-2*molxx283.89*g*mol^-1=??g$

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In the reaction $P_{4} + 5 O_{2} \to P O_{4} O_{10}$ $P_4 + 5O_2 -> PO_4O_10$ , 7.75 g $P_{4}$ $P_4$ reacts. How do you calculate the mass of $P_{4} O_{10}$ $P_4O_10$ produced?

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In the reaction P4+5O2→PO4O10P_4 + 5O_2 -> PO_4O_10, 7.75 g P4P_4 reacts. How do you calculate the mass of P4O10P_4O_10 produced?

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Explanation:

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In the reaction $P_{4} + 5 O_{2} \to P O_{4} O_{10}$ $P_4 + 5O_2 -> PO_4O_10$ , 7.75 g $P_{4}$ $P_4$ reacts. How do you calculate the mass of $P_{4} O_{10}$ $P_4O_10$ produced?