# Mole Calculations - I have answered some but not sure if I am doing it right. Could you please tell me if I am doing it right if not correct me so far and the other ones on the sheet too?

Jan 29, 2017

Please Note: you are to divide mass by molar mass to calculate the number of interacting moles and finally to equate the ratio of interacting moles calculated to the corresponding mole ratio obtained from the balanced equation

Q-1)
Balanced Equation

$2 A {l}_{2} {O}_{3} \text{ "->" "4Al" } + 3 {O}_{2}$

Molar mass
$g \text{/mol }$ (227+316)=102 $\text{ "" }$ 27

Mass in gram " "1020" "" "" "x=?

Mole ratio $\text{ "" "2" "" "" "" "" "" } 4$
(as per equation)

So $\frac{\frac{x}{27}}{\frac{1020}{102}} = \frac{4}{2}$

$\implies x = 10 \times 2 \times 27 = 540 g$

Q-2)

Balanced Equation

$2 C a \text{ "+" "O_2" } \to 2 C a O$

Molar mass
$g \text{/mol "" "" }$ 40 $\text{ "" "" }$ 2x16=32

Mass in gram " "10" "" "" "" "x=?

Mole ratio $\text{ "" "2" "" "" "" "" "" } 1$
(as per equation)

So $\frac{\frac{x}{32}}{\frac{10}{40}} = \frac{1}{2}$

$\implies x = 4 g$

Q-3)

Balanced Equation

${C}_{3} {H}_{8} \text{ } + 5 {O}_{2} \to 3 C {O}_{2} + 5 {H}_{2} O$

Molar mass
$g \text{/mol }$3x12+8x1=44 $\text{ }$ 2x16=32

Mass in gram $\text{ "x=?" "" "" } 50$

Mole ratio $\text{ "" "1" "" "" "" "" "" } 5$
(as per equation)

So $\frac{\frac{x}{44}}{\frac{50}{32}} = \frac{1}{5}$

$\implies x = \frac{50 \times 44}{5 \times 32} g = 13.75 g$

Q-4)

Balanced Equation

$3 {H}_{2} \text{ "+" "N_2" } \to 2 N {H}_{3}$

Molar mass
$g \text{/mol "" }$ 2x1=2 $\text{ "" }$ 2x14=28 $\text{ }$2(14+3x1)=34

Mass in gram " "20" "" "" "" "" "" "" "" "x=?

Mole ratio $\text{ "" "3" "" "" "" "" "" "" "" "" } 2$
(as per equation)

So $\frac{\frac{x}{34}}{\frac{20}{2}} = \frac{2}{3}$

$\implies x = \frac{20}{3} \times 34 g = 226.67 g$

Q-5)

Balanced Equation

${H}_{2} S {O}_{4} + \text{ } 2 N a O H \to N {a}_{2} S {O}_{4} + 2 {H}_{2} O$

Molar mass $\text{ }$ 2x1+32+4x16 $\text{ }$ (23+16+1)
$g \text{/mol "" "" "=98" "" "" "" } = 40$

Mass (in kg)" "" "10" "" "" "" "x=?

Mole ratio $\text{ "" "1" "" "" "" "" "" } 2$
(as per equation)

So $\frac{\frac{x}{40} \times {10}^{3}}{\frac{10}{98} \times {10}^{3}} = \frac{2}{1}$

$\implies x = \frac{10 \times 2 \times 40}{98} k g \approx 8.163 k g$

Q-6)

Balanced Equation

$C u C {O}_{3} \text{ "->CuO" } + C {O}_{2}$

Molar mass $\text{ }$ 63.5+12+3x16 $\text{ "" "" }$ (12+2x16)
$g \text{/mol "" "" "=123.5" "" "" "" "" "" } = 44$

Mass in gram " "10" "" "" "" "" "" "" "" "x=?

Mole ratio $\text{ "" "1" "" "" "" "" "" "" "" "" } 1$
(as per equation)

So $\frac{\frac{x}{44}}{\frac{10}{123.5}} = \frac{1}{1}$

$\implies x = \frac{440}{123.5} = 3.56 g$