Question

If you have 560 grams of magnesium. what mass of magnesium oxide will be produced?

Answer 1

Here's what I get.

Step-by-step explanation:

`\begin{array}{lcccc}\textbf{(a)} \qquad \text{Na} & \text{reacts with} & \text{ diatomic chlorine}& \text{to form} & \text{sodium chloride}\\\textbf{(b)}\qquad \text{Na} & + & \text{Cl}_(2) & \longrightarrow \, & \text{NaCl}\\\textbf{(c)}\qquad \text{2Na}& + & \text{Cl}_(2) & \longrightarrow \, & \text{2NaCl}\\\end{array}`

(d) Mass of chlorine

(i) We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 70.91 58.44

2Na + Cl₂ ⟶ 2NaCl

m/g: 117

(ii) Calculate the moles of NaCl

`\text{Moles of NaCl} = \text{117 g NaCl} * \frac{\text{1 mol NaCl}}{\text{58.44 g NaCl }} = \text{2.002 mol NaCl }`

(iii) Calculate the moles of Cl₂

The molar ratio is (1 mol Cl₂ /2 mol NaCl)

`\text{Moles of Cl$_(2)$}= \text{2.002 mol NaCl} * \frac{\text{1 mol Cl$_(2)$}}{\text{2 mol NaCl}} = \text{1.001 mol Cl$_(2)$}`

(iv) Calculate the mass of Cl₂

`\text{Mass of Cl$_(2)$} = \text{1.001 mol Cl$_(2)$} * \frac{\text{70.91 g Cl$_(2)$}}{\text{1 mol Cl$_(2)$}} = \text{71.0 g Cl$_(2)$}\\\\\text{The mass of chlorine needed is } \boxed{\textbf{71.0 g Cl$_(2)$}}`

Answer 2

=933.3 grams

Step-by-step explanation:

Magnesium reacts with oxygen producing magnesium oxide according to the following equation.

2Mg₍s₎ + O₂₍g₎→ 2MgO₍s₎

From the above equation, 2 moles of Magnesium produces 2 moles of magnesium oxide. Therefore the ratio of magnesium to magnesium oxide produced is 1:1

1 mole of magnesium oxide has a mass of: 24+16 =40 grams

560 grams of magnesium=560/24 moles

=70/3 moles.

Since the ratio is 1:1 the reaction produces 70/3 moles of magnesium oxide.

70/3 moles of MgO= 70/3 moles× 40 grams

=933.3 grams