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dem82 [27]
3 years ago
14

The reaction of hydrogen gas and iron oxide is described by the chemical equation below. 3H2+Fe2O3→2Fe+3H2O How many moles of ir

on will be produced if 3.3mol of hydrogen gas reacts with 1.5mol of iron oxide?
Chemistry
1 answer:
ELEN [110]3 years ago
5 0

Answer:

2.2 moles of Fe will be produced

Explanation:

Step 1: Data given

Number of moles of hydrogen gas = 3.3 moles

Number of moles of iron oxide = 1.5 moles

Step 2: The balanced equation

3H2 + Fe2O3 → 2Fe + 3H2O

Step 3: Calculate the limiting reactant

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

Hydrogen gas is the limiting reactant. It will completely be consumed (3.3 moles). Fe2O3 is in excess. There will react 3.3 / 3 = 1.1 moles

There will remain 1.5 - 1.1 = 0.4 moles Fe2O3

Step 4: Calculate moles Fe

For 3 moles H2 we need 1 mol Fe2O3 to produce 2 moles Fe and 3 moles H2O

For 3.3 moles H2 we'll have 2/3 * 3.3 = 2.2 moles Fe

2.2 moles of Fe will be produced

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Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?
blagie [28]
<h2>Answer:</h2>

She should notify her teacher.

<h2>Explanation:</h2>

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4 0
3 years ago
The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
son4ous [18]

Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

5 0
3 years ago
A gas has a volume of 3L at 200 kPa. What will its volume be if the pressure is changed to 500 kPa?
marta [7]

Answer:

6L

Explanation:

<em>if it's 3L per 200kPa</em>

then it would be;

4L per 300kPa

5L per 400kPa

6L per 500kPa

that's how i'd work it out in my head, hope it helps, but not sure though!

5 0
3 years ago
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