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Alex_Xolod [135]
3 years ago
12

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N 2 (g) 3H 2 (g) 2NH3 (g) I

f the reaction yield is 87.5%, how many moles of N 2 are needed to produce 3.00 mol of NH 3
Chemistry
1 answer:
balandron [24]3 years ago
5 0

Answer:

We need 1.714 moles N2

Explanation:

Step 1: Data given

The reaction yield = 87.5 %

Number of moles NH3 produced = 3.00 moles

Step 2: The balanced equation

N2(g)+ 3H2(g) →2NH3(g)

Step 3: Calculate moles N2

For 2 moles NH3 produced we need 1 mol N2 and 3 moles H2

This means, if the yield was 100%, for 3.00 moles NH3 produced , we need 1.5 moles N2

For a 87.5 % yield:  

we need more N2, increased by a ratio of 100/87.5.

100/87.5 * 1.5 =  1.714 moles N2

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What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4
Ivenika [448]

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130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

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(M·V)concentrate = (M·V)diluted

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Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

5 0
3 years ago
Can anybody check my answer?
anzhelika [568]

Answer:

\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

25. Boyle's Law

The temperature and amount of gas are constant, so we can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Data:

\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

Calculations:

\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

26. Ideal Gas Law

We have p, V and n, so we can use the Ideal Gas Law to calculate the volume.

pV = nRT

Data:  

p = 101.3 kPa

V = 20 L

n = 5 mol

R = 8.314 kPa·L·K⁻¹mol⁻¹

Calculation:

101.3 × 20 = 5 ×  8.314 × T

2026 = 41.57T

T = \dfrac{2026}{41.57} = \textbf{49 K}\\\\\text{The Kelvin temperature is }\boxed{\textbf{49 K}}

6 0
3 years ago
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