Answer to Question #158611 in General Chemistry for Someone

Question

Calculate reacting masses from experimental data questions

1. In the complete combustion of ethene, 22g of carbon dioxide is produced. What mass

(in grams) of oxygen is used?

ethene + oxygen → carbon dioxide + water

C₂H₄ + 3 O₂ → 2 CO₂ + 2 H₂O

2. In the displacement reaction between chlorine and potassium bromide, 320g of

bromine is produced. What mass (in grams) of chlorine is used?

3. In the neutralisation reaction between sulfuric acid and copper (II) carbonate, 159.5g

of copper (II) sulfate is produced. What mass (in grams) of carbon dioxide is also

produced?

4. In the neutralisation reaction between hydrochloric acid and sodium hydroxide,

29.25g of sodium chloride is produced. What mass (in grams) of water is also

produced?

5. In the combustion of hydrogen, 80g of oxygen is used. What mass (in grams) of

hydrogen is also used?

hydrogen + oxygen → water

Answer 1

Solution:

1)

ethene + oxygen → carbon dioxide + water

C₂H₄ + 3O₂ → 2CO₂ + 2 H₂O 

Molar mass of CO₂ = 12 + 16×2 = 44 $(\frac{g}{mole})$

Molar mass of O₂ = 16 × 2 = 32 $(\frac{g}{mole})$

According to the equation, mole ratio O₂ : CO₂ = 3 : 2

Moles of CO₂ produced = $\dfrac{(22g)}{(44(\frac{g}{mole}))}$ = 0.5 mol

Moles of O₂ produced = (0.5 mol) × ($\frac{3}{2}$ ) = 0.75 mol

Mass of O₂ produced = (0.75 mol) × (32 $(\frac{g}{mole})$) = 24 g

2)

2KBr  +  Cl2  $→$   2KCl  +  Br2

Br₂  =160 $(\frac{g}{mole})$

Cl₂  = 71 $(\frac{g}{mole})$

mass of chlorine = $\dfrac{320g}{160(\frac{g}{mole} )}*71 (\frac{g}{mole})=142 g$

3)

Copper II Carbonate + Sulfuric acid →Copper Sulfate + Water + Carbon Dioxide

CuCO₃ + H₂SO₄ → CuSO₄ + H₂O + CO₂

123.6 g of CuCO₃ + 98 g of H₂SO₄ →159.6 g of CuCO₃ + 18 g of H₂O + 44 g of CO₂

check: 123.6 g+ 98 g = 159.6 + 18 + 44

H₂SO₄ = 2+32.1+64 = 98 $(\frac{g}{mole})$

CuSO₄ = 63.6+32+64 = 159.6 $(\frac{g}{mole})$

CuCO₃ = 123.6 $(\frac{g}{mole})$

Water = 18 $(\frac{g}{mole})$

CO₂ = 44 $(\frac{g}{mole})$

ratio of CO₂ to CuCO₃ is $\tfrac{44}{159.6}$

$\tfrac{44}{159.6}$ =$\tfrac{x}{159.6}$

x = 44

4)

Molar mass of NaCl = (23.0 + 35.5) $(\frac{g}{mole})$ = 58.5 $(\frac{g}{mole})$

Molar mass of H₂O = (1.0×2 + 16.0) $(\frac{g}{mole})$ = 18.0 $(\frac{g}{mole})$

Equation for the reaction:

HCl + NaOH → NaCl + H₂O

Mole ratio  NaCl : H₂O = 1 : 1

Moles of NaCl produced =$\dfrac{29.25g}{58.5(\frac{g}{mole})}$ = 0.5 mol

Moles of H₂O produced = 0.5 mol

Mass of H₂O produced = (0.5 mol) × (18.0 $(\frac{g}{mole})$) = 9.00 g

5)

Hydrogen + Oxygen ➜ water

2H₂(g) + O₂(g) ➜ 2H₂O(l)

molecular weights

H = 1.008

O = 15.999

2H₂ = 4.032

O₂ = 31.998

2H₂O = 36.03

2 mole of H₂ + 1 moles of O₂ ➜ 2 moles of H₂O

4.03 grams of H₂ + 32 grams of O₂ ➜ 36.03 grams of H₂O

80g of oxygen is used. What mass (in grams) of hydrogen has also used

the ratio of H₂ to O₂ is $\tfrac{4.03}{32}$

$\tfrac{4.03}{32}$ = $\tfrac{x}{80}$

x = 80•$\tfrac{4.03}{32}$ = 10.1 g