Question

Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
3) Convert 0.250 moles of aluminum sulfate to grams.

4) Convert 2.70 grams of ammonia to moles.

Answer 1

Answer:

0.000731 grams aluminium sulfate

46.0 mols ammonia

Explanation:

ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol

$ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g$

NH3 has a molar mass of 17.031 g/mol

$NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols$

Answer 2

Aluminium sulphate (AlS) whose molar mass is= $\sf{ 342.15\dfrac{g}{mol} }$

we have to find the 0.250 moles of aluminum sulphate.

$\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}$

Ammonia(NH3) whose molar mass is =$\sf{17.031\dfrac{mol}{g} }$

We have to find 2.70 grams of ammonia

$\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}$