Question

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

Answer 1

Answer:

=933.3 grams

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

Answer 2

Answer:

Here's what I get.

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

$\text{Moles of NaCl} = \text{117 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl }} = \text{2.002 mol NaCl }$

(iii) Calculate the moles of Cl₂

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

$\text{Moles of Cl _{2} }= \text{2.002 mol NaCl} \times \dfrac{\text{1 mol Cl _{2} }}{\text{2 mol NaCl}} = \text{1.001 mol Cl _{2} }$

(iv) Calculate the mass of Cl₂

$\text{Mass of Cl _{2} } = \text{1.001 mol Cl _{2} } \times \dfrac{\text{70.91 g Cl _{2} }}{\text{1 mol Cl _{2} }} = \text{71.0 g Cl _{2} }\\\\\text{The mass of chlorine needed is } \boxed{\textbf{71.0 g Cl _{2} }}$