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3 years ago

How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?

Chemistry

1 answer:

Kitty [74]3 years ago

3 0

Answer:

21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C7H14 + 21O2 —> 14CO2 + 14H2O

From the balanced equation above, 2L of C7H14 produced 14L of H2O.

Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.

Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2O will occupy

26.25L i.e

Xmol of H2O = 26.25/22.4

Xmol of H2O = 1.172 mole

Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

Number of mole H2O = 1.172 mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O =..?

Mass = mole x molar mass

Mass of H2O = 1.172 x 18

Mass of H2O = 21.10g

Therefore, 21.10g of H2O is produced from the reaction.

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Answer:

$\boxed {\boxed {\sf B. \ 1170 \ Joules }}$

Explanation:

We are asked to find how much heat a sample of copper absorbs when the temperature is increased.

Since we know the mass, temperature increase, and specific heat capacity, we can use the following formula to calculate heat.

$q= mc \Delta T$

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m= 100 g
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Substitute the values into the formula.

$q= (100 \ g )(0.39 \ J/g \textdegree C ) (30.0 \textdegree C )$

Multiply the first two values. Note that the units of grams cancel.

$q= 39 \ J/ \textdegree C (30.0 \textdegree C )$

Multiply again, this time the units of degrees Celsius cancel.

$q= 1170 \ J$

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