District of Springfield MO, R-12

Question 1 (2 pts)

The balanced equation [tex]P_4(s) + 6 H_2(g) \rightarrow 4 PH_3(g)[/tex] tells us that 5.0 mol of [tex]\(H_2\)[/tex] reacts with 2.5 mol of [tex]\(P_4\)[/tex].



Answer :

Sure, let's solve this step-by-step.

We start with the balanced chemical equation:

[tex]\[ P_4(s) + 6 H_2(g) \rightarrow 4 PH_3(g) \][/tex]

This equation tells us how the reactants combine and form products. According to the equation:

- 1 mole of [tex]\( P_4 \)[/tex] reacts with 6 moles of [tex]\( H_2 \)[/tex] to produce 4 moles of [tex]\( PH_3 \)[/tex].

Now, we are given that we have 2.5 moles of [tex]\( P_4 \)[/tex]. We need to determine how many moles of [tex]\( H_2 \)[/tex] are required to completely react with 2.5 moles of [tex]\( P_4 \)[/tex].

To do this, we use the molar ratio from the balanced equation. The ratio between [tex]\( P_4 \)[/tex] and [tex]\( H_2 \)[/tex] in the reaction is 1:6. This means:

[tex]\[ 1 \, \text{mol} \, P_4 \text{ reacts with} 6 \, \text{mol} \, H_2 \][/tex]

To find out how many moles of [tex]\( H_2 \)[/tex] are needed for 2.5 moles of [tex]\( P_4 \)[/tex], we set up a proportion based on the molar ratio:

[tex]\[ \frac{1 \, \text{mol} \, P_4}{6 \, \text{mol} \, H_2} = \frac{2.5 \, \text{mol} \, P_4}{x \, \text{mol} \, H_2} \][/tex]

Here, [tex]\( x \)[/tex] is the number of moles of [tex]\( H_2 \)[/tex] required. We solve for [tex]\( x \)[/tex]:

[tex]\[ x = 2.5 \, \text{mol} \, P_4 \times \frac{6 \, \text{mol} \, H_2}{1 \, \text{mol} \, P_4} \][/tex]

[tex]\[ x = 2.5 \times 6 \][/tex]

[tex]\[ x = 15 \][/tex]

So, to react completely with 2.5 moles of [tex]\( P_4 \)[/tex], you need 15 moles of [tex]\( H_2 \)[/tex].