Question

do you remember how to convert mass and moles?
on the next screen, youre going to look for the limiting reactant in a chemical reaction. before you do that, you need to make sure youre able to convert mass and moles using molar mass.
solve the problems below to be sure youre ready to proceed.
5 sf
345.50 g of $c_{6}h_{12}o_{6}=$mol $c_{6}h_{12}o_{6}$ (mm = 180.16 g/mol)
4 sf
100.0 g of $o_{2}=$mol $o_{2}$ (mm = 32.00 g/mol)
4 sf
14.37 moles of $agno_{3}=$g $agno_{3}$ (mm = 169.87 g/mol)

Explanation:

Step1: Convert mass of \(C_6H_{12}O_6\) to moles

Use the formula \(n=\frac{m}{M}\), where \(n\) is moles, \(m\) is mass, and \(M\) is molar mass.
For \(C_6H_{12}O_6\), \(m = 345.50\ g\) and \(M=180.16\ g/mol\).
\(n=\frac{345.50}{180.16}\approx1.9177\ mol\)

Step2: Convert mass of \(O_2\) to moles

For \(O_2\), \(m = 100.0\ g\) and \(M = 32.00\ g/mol\).
Using \(n=\frac{m}{M}\), \(n=\frac{100.0}{32.00}=3.125\ mol\)

Step3: Convert moles of \(AgNO_3\) to mass

Use the formula \(m=n\times M\), where \(n = 14.37\ mol\) and \(M=169.87\ g/mol\) for \(AgNO_3\).
\(m=14.37\times169.87 = 14.37\times(170 - 0.13)=14.37\times170-14.37\times0.13=2442.9 - 1.8681=2441.0319\approx2441.0\ g\)

Answer:

\(1.9177\ mol\ C_6H_{12}O_6\), \(3.125\ mol\ O_2\), \(2441.0\ g\ AgNO_3\)