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ale4655 [162]
3 years ago
14

How many grams of water can be formed from the reaction of 8.76 grams of H2 with 10.5 liters of O2 (at STP) according to the bal

anced equation: 2 H2 + O2 --> 2 H2O Express your answer to 3 sig figs. Do NOT include units!
Chemistry
1 answer:
Kisachek [45]3 years ago
8 0

Answer:

16.9g of H₂O can be formed

Explanation:

Based on the chemical reaction, 2 moles of H₂ react per mole of O₂. To anser this question we must find limiting reactant converting the mass and volume of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

8.76g * (1mol / 2.016g) = 4.345 moles

<em>Moles O₂:</em>

PV = nRT

PV/RT = n

P = 1atm at STP

V = 10.5L

R = 0.082atmL/molK

T = 273.15K at STP

n = 1atm*10.5L / 0.082atmL/molK*273.15K

n = 0.469 moles of oxygen

For a complete reaction of 4.345 moles moles of hydrogen are required:

4.345 moles H2 * (1mol O2 / 2mol H2) = 2.173 moles of O2 are required. As there are just 0.469 moles, Oxygen is limiting reactant

Now, 1 mole of O2 produce 2 moles of H2O. 0.469 moles will produce:

0.469 moles O₂ * (2 moles H₂O / 1mol O₂) = 0.938 moles H₂O.

The mass is -Molar mas H₂O = 18.01g/mol-:

0.938 moles * (18.01g/mol) =

<h3>16.9g of H₂O can be formed</h3>
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4. How many grams of ammonium carbonate are needed to decompose in order to produce
Thepotemich [5.8K]

Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

(NH4)2CO3 –> (NH4)2O + CO2

Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

= 2[18] + 60 = 96g/mol

Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

4 0
3 years ago
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