Method that used to separate kerosine and water?

Chemistry

2 answers:

nata0808 [166]4 years ago

7 0

You can use a separating funnel. The person will go the top and the water to the bottom.

earnstyle [38]4 years ago

6 0

Basically you could just use a separating funnel

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What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3?

Anvisha [2.4K]

Answer:

$\boxed{\text{53.3 \%}}$

Explanation:

MM: 2.016 17.03

N₂ + 3H₂ ⟶ 2NH3

m/g: 26.3

1. Theoretical yield

(a) Moles of H₂

$\text{Moles of H${_2}$} = \text{26.3 g H${_2}$} \times \dfrac{\text{1 mol H${_2}$}}{\text{2.016 g H${_2}$}} = \text{13.05 mol H${_2}$}$

(b) Moles of NH₃

$\text{Moles of NH${_3}$} = \text{13.05 mol H${_2}$} \times \dfrac{\text{2 mol NH${_3}$}}{\text{3 mol H${_2}$}} = \text{8.697 mol NH${_3}$}$

(c) Theoretical yield of NH₃

$\text{Mass of NH${_3}$} = \text{8.967 mol NH${_3}$} \times \dfrac{\text{17.03 g NH${_3}$}}{\text{1 mol NH${_3}$}} = \text{148.1 g NH${_3}$}$

(d) Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \% = \dfrac{\text{79.0 g}}{\text{148.1 g}} \times 100 \% = \textbf{53.3 \%}\\\\\text{The percent yield is }\boxed{\textbf{53.3 \%}}$