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1 year ago

For the reaction, calculate how many moles of the product form when 0.046 mol of O2 completely reacts.

Chemistry

1 answer:

aleksandrvk [35]1 year ago

3 0

Answer: 0.023 moles of CaO.

Explanation:

The balanced equation tells us that we'll get 2 moles of CaO for every one mole of O2. Since we have 0.046 moles of O2, we'll get 0.023 moles of CaO. (2 sig figs)

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43 mg = [?]g <br>A. 0.043 g <br>B. 4.3 g <br>C. 4300 g <br>D. 43,000 g

Ymorist [56]

Answer:

Option A (0.043 g) is the correct answer.

Explanation:

Given:

= 43 mg

As we know,

$1 \ mg = \frac{1}{1000} \ g$

then,

⇒ $43 \ mg = \frac{43}{1000} \ g$

$= 0.043 \ g$

Thus, the above is the correct alternative.

4 0

2 years ago

A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release

Marrrta [24]

Answer:

$V_2=12.1L$

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

$T_1=20+273=293K\\\\T_2=-40+273=233K$

Then, we obtain:

$V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L$

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5 0

2 years ago

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Nuetrik [128]

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$ ⇒ $2ScCl_3$

The above reaction shows that 2 moles of Sc can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$