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Chapter 3: Problem 15

How many moles of calcium (Ca) atoms are in \(77.4 \mathrm{~g}\) of \(\mathrm{Ca} ?\)

Short Answer

Expert verified
There are approximately \(1.93\) moles of calcium (Ca) atoms in \(77.4 \mathrm{~g}\) of Ca.

Step by step solution

01

Identify Given Values

The problem provides the mass of calcium, which is \(77.4 \mathrm{~g}\). The molar mass of calcium, found on the periodic table, is approximately \(40.08 \mathrm{~g/mol}\). We will use these two values in our calculation.
02

Calculate the Number of Moles

The formula to calculate the number of moles is \(\text{Moles} = \frac{\text{Mass(g)}}{\text{Molar mass(g/mol)}}\). Plugging in our given values, we find \(\text{Moles} = \frac{77.4 \mathrm{~g}}{40.08 \mathrm{~g/mol}}\).
03

Simplify the Calculation

When you divide \(77.4 \mathrm{~g}\) by \(40.08 \mathrm{~g/mol}\), the g unit in the numerator and the denominator will cancel out, leaving you with the answer in the unit of moles (mol).
04

Round the Answer

After performing the division, the number of moles will be rounded to three significant figures since the number with the least significant figures in the problem is three (77.4 g). It's crucial to carry the rounding process correctly.

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Most popular questions from this chapter

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition can be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

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