Question

Given the following reaction:

Answer 1

15.76 grams of AgCl  will be produced from 7.00 g of NaCl and 95.0 g of AgNO3.

Explanation:

Balanced equation for the reaction:

NaCl + AgNO3 → AgCl + NaNO3

given:

mass of NaCl = 7  gram

mass of AgNO3 = 90 GRAM

atomic mass of NaCl = 58.44 grams/mole

atomic mass of AgNO3 = 169.87 grams/mole

number of moles = $\frac{mass}{atomic mass of onemole}$    1 equation

putting the values in equation 1

number of moles of NaCl = $\frac{7}{58.44}$

                                            = 0.11 moles

number of moles of AgNO3 = $\frac{95}{169.87}$

                                               = 0.55

the limiting reagent is NaCl as it is a reactant that produces small quantity of AgCl

1 Mole of NaCl reacted to form 1 mole of AgCl

0.11 mole of NaCl will produce x moles of AgCl

$\frac{1}{1}$ = $\frac{x}{0.11}$

0.11 moles of AgCl is produced

atomic mass of AgCl = 143.32

mass = 0.11 x 143.32

          = 15.76 grams of AgCl is produced.

Answer 2

Answer:

17.2

Explanation: