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# Consider the reaction:P4(s) + 6H2(g) ? 4 PH3(g)(a) If 88.6 ## Problem 114P Chapter 11

Introductory Chemistry | 5th Edition

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Problem 114P

PROBLEM 114P

Consider the reaction:

P4(s) + 6H2(g) → 4 PH3(g)

(a) If 88.6 L of H2(g), measured at STP, is allowed to react with 158.3 g of P4, what is the limiting reactant?

(b) If 48.3 L of PH3, measured at STP, forms, what is the percent yield?

Step-by-Step Solution:

Solution 114P

Step 1:

Given reaction:

P4(s) + 6H2(g) → 4PH3(g)

(a) If 88.6 L of H2(g), measured at STP, is allowed to react with 158.3 g of P4, what is the limiting reactant?

To find the limiting reactant, we should know the moles of each reactant. So let’s find the number of moles of each reactant :

First, let’s find number of moles of H2 :

We know, at STP(Standard Temperature and Pressure), one mole of any gas occupies 22.4 L.

Moles of H2 = = 3.955 moles.

Now, let’s find number of moles of P4 :

Molar mass of P4 = 123.89 g/mol.

moles of P4 = = 1.277 moles.

Step 2:

From the given reaction, we see that 6 moles of H2 react with 1 mole of P4, i.e the ratio of reactants P4 to H2 is :

1 : 6

Hence, number of moles of H2 required is :

= 1.277 moles of P4 6

= 7.66 moles of H2 required.

Since there is 3.955 moles of H2, H2 is limiting reactant

Step 3 of 4

Step 4 of 4

##### ISBN: 9780321910295

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Consider the reaction:P4(s) + 6H2(g) ? 4 PH3(g)(a) If 88.6

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