Question

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

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 of \(Na_2SO_4\)

Molar mass of \(Na = 22.99\ g/mol\), \(S = 32.07\ g/mol\), \(O = 16.00\ g/mol\)
\(M(Na_2SO_4)=2\times22.99 + 32.07+4\times16.00=142.05\ g/mol\)
\(\frac{30\ g}{1}\times\frac{1\ mol}{142.05\ g}\approx0.211\ mol\)

Step2: Calculate molar mass of \(BeCl_2\)

Molar mass of \(Be = 9.01\ g/mol\), \(Cl = 35.45\ g/mol\)
\(M(BeCl_2)=9.01 + 2\times35.45=79.91\ g/mol\)
\(\frac{25\ g}{1}\times\frac{1\ mol}{79.91\ g}\approx0.313\ mol\)

Step3: Calculate molar mass of \(Fe_3(PO_4)_2\)

Molar mass of \(Fe = 55.85\ g/mol\), \(P = 30.97\ g/mol\), \(O = 16.00\ g/mol\)
\(M(Fe_3(PO_4)_2)=3\times55.85+2\times(30.97 + 4\times16.00)=357.49\ g/mol\)
\(\frac{110\ g}{1}\times\frac{1\ mol}{357.49\ g}\approx0.308\ mol\)

Step4: Calculate molar mass of \(CuNO_3\)

Molar mass of \(Cu = 63.55\ g/mol\), \(N = 14.01\ g/mol\), \(O = 16.00\ g/mol\)
\(M(CuNO_3)=63.55+14.01+3\times16.00=125.56\ g/mol\)
\(\frac{1.1\ g}{1}\times\frac{1\ mol}{125.56\ g}\approx0.00876\ mol\)

Step5: Calculate moles of \(SiO_2\)

Given \(M(SiO_2) = 60.09\ g/mol\)
\(\frac{45\ g}{1}\times\frac{1\ mol}{60.09\ g}\approx0.749\ mol\)

Answer:

1. \(0.211\ mol\)
2. \(0.313\ mol\)
3. \(0.308\ mol\)
4. \(0.00876\ mol\)
5. \(0.749\ mol\)