Question

how many moles of br₂ are required to react with 2.22 x 10² moles of n₂? (answer = 6.66 x 10² mol br₂) note: the reaction must be balanced, dont leave anything blank. 1 n₂ + 2 br₂ → 3 nbr₃ + 44.0 kj 4 mol br₂ (2.22 x 10² mol n₂)(-------------) = 5 mol br₂ 6 mol n₂ a. 1 b. 2 c. 3 d. 4 e. 5 f. 253.719 g. 3.38 x 10⁰ h. 3.26 x 10³ i. 44.0 j. 1.33 x 10⁻² k. 9.77 x 10³ l. 4.44 x 10² m. 6.66 x 10² n. 500.0 o. 556 p. 90.0 q. 3.2852 r. 159.81 s. 28.013 t. 1.750 u. 0.8750 v. 13.0 w. 0.46306 x. 1.3892 y. 24.5 z. 38.5 aa. 2.190 bb. 250.

Explanation:

Step1: Balance the chemical equation

The balanced chemical equation is \(N_{2}+3Br_{2}\to 2NBr_{3}+44.0\ kJ\). So the coefficients are \(1\) (for \(N_{2}\)), \(3\) (for \(Br_{2}\)), \(2\) (for \(NBr_{3}\)).

Step2: Set up the mole - ratio calculation

From the balanced equation, the mole ratio of \(Br_{2}\) to \(N_{2}\) is \(\frac{3\ mol\ Br_{2}}{1\ mol\ N_{2}}\).
We have \(n(N_{2}) = 2.22\times10^{2}\ mol\). Using the formula \(n(Br_{2})=n(N_{2})\times\frac{\text{moles of }Br_{2}}{\text{moles of }N_{2}}\)

$$n(Br_{2})=(2.22\times 10^{2}\ mol\ N_{2})\times\frac{3\ mol\ Br_{2}}{1\ mol\ N_{2}}$$

$$n(Br_{2})=(2.22\times3)\times10^{2}\ mol\ Br_{2}$$

$$n(Br_{2}) = 6.66\times10^{2}\ mol\ Br_{2}$$

Answer:

1. A. \(1\) (for the coefficient of \(N_{2}\) in the balanced equation)
2. C. \(3\) (for the coefficient of \(Br_{2}\) in the balanced equation)
3. B. \(2\) (for the coefficient of \(NBr_{3}\) in the balanced equation)
4. C. \(3\) (the numerator of the mole - ratio \(\frac{3\ mol\ Br_{2}}{1\ mol\ N_{2}}\))
5. M. \(6.66\times 10^{2}\) (the moles of \(Br_{2}\) calculated)
6. A. \(1\) (the denominator of the mole - ratio \(\frac{3\ mol\ Br_{2}}{1\ mol\ N_{2}}\))