Question

part ii: converting grams to moles - using the mw from part i, determine the number of moles in the following.

1. \\( \frac { 30 \text { grams of } \mathrm { na } _ { 2 } \mathrm { so } _ { 4 } } { 1 } \times \\)
2. \\( \frac { 25 \text { grams of } \mathrm { becl } _ { 2 } } { 1 } \times \\)
3. \\( \frac { 110 \text { grams of } \mathrm { fe } _ { 3 } ( \mathrm { po } _ { 4 } ) _ { 2 } } { 1 } \times \\)
4. \\( \frac { 1.1 \text { grams of } \mathrm { cuno } _ { 3 } } { 1 } \times \\)
5. \\( \frac { 45 \text { grams of } \mathrm { sio } _ { 2 } } { 1 } \times \\)

Explanation:

Step1: Calculate molar mass (MW)

• For \(Na_2SO_4\): \(MW=(2\times23)+32+(4\times16)=142\space g/mol\)
• For \(BeCl_2\): \(MW = 9+(2\times35.5)=80\space g/mol\)
• For \(Fe_3(PO_4)_2\): \(MW=(3\times56)+2\times(31+(4\times16)) = 358\space g/mol\)
• For \(CuNO_3\): \(MW = 64+14+(3\times16)=126\space g/mol\)
• For \(SiO_2\): \(MW = 28+(2\times16)=60\space g/mol\)

Step2: Use formula \(n=\frac{m}{MW}\)

• 1. \(\frac{30\space g}{1}\times\frac{1\space mol}{142\space g}\approx0.211\space mol\)
• 2. \(\frac{25\space g}{1}\times\frac{1\space mol}{80\space g}= 0.3125\space mol\)
• 3. \(\frac{110\space g}{1}\times\frac{1\space mol}{358\space g}\approx0.307\space mol\)
• 4. \(\frac{1.1\space g}{1}\times\frac{1\space mol}{126\space g}\approx0.00873\space mol\)
• 5. \(\frac{45\space g}{1}\times\frac{1\space mol}{60\space g}=0.75\space mol\)

Answer:

1. \(0.211\space mol\)
2. \(0.3125\space mol\)
3. \(0.307\space mol\)
4. \(0.00873\space mol\)
5. \(0.75\space mol\)