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3 years ago

How many grams of water are made from the reaction of 16.0 grams of oxygen gas?

1 answer:

4 0

Assuming hydrogen gas is in excess, calculate the moles of oxygen

n(O2) = 16.0 g / 32.0 g/mol = 0.500 mol 

Calculate the moles of water using the mole ratio; the ratio of H2O to O2 is 2:1 or the moles of H2O is twice the moles of O2 

n(H2O) = 2 x n(O2) = (2)(0.500) = 1.00 mol 

Calculate the mass of H2O by multiplying the moles by the molar mass of water 

mass = (1.00 mol)(18.0 g/mol) = 18.0 g H2O 

Ans: A)

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Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

Answer:

For A: The answer becomes 0.1

For B: The answer becomes 0.4884

Explanation:

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

Rule applied for addition and subtraction:

The least precise number present after the decimal point determines the number of significant figures in the answer.

Rule applied for multiplication and division:

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

For A: (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

For B: (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

6 0

3 years ago

How many atoms are in .35 mol of water?

Tom [10]

2.91 is the amount of atoms

6 0

2 years ago

6.0 mol Al reacts with 4.0 mol O2 to form Al2O3.

Tema [17]

Answer:

3.0 moles Al₂O₃

Explanation:

We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.

4 Al + 3 O₂ -----> 2 Al₂O₃

6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al

4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂

As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.

However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.

6 0

1 year ago

What happens to glucose in the Calvin cycle? Plz help

Leona [35]

The reactions of the Calvin cycle add carbon from carbon dioxide in the atmosphere to a five-carbon molecule known as RuBP. These reactions use chemical energy that were produced in the light reactions, from NADPH and ATP. The final product of the Calvin cycle is glucose.

7 0

3 years ago

How many microliters are in 7.4 x 10^-61 centimeters?

schepotkina [342]

Answer:

Microliters = $7.4\times 10^{-58}$