Answer
a. 26.6 g of $K_2CO_3$
b. 4.46 g of $AgNO_3$
c. 16.7 g of $H_3PO_4$
Work Step by Step
a.
1. Find the conversion factors
- 0.450 M $K_2CO_3$ solution: $$0.450 \space mole \space K_2CO_3 = 1 \space L \space solution$$
- Molar mass: $ K_2CO_3 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )+ ( 39.10 $\times$ 2 )= 138.21 g/mol
2. Calculate the mass in grams:
$$0.428 \space L \space solution \times \frac{0.450 \space mole \space K_2CO_3}{1 \space L \space solution} \times \frac{138.21 \space g \space K_2CO_3}{1 \space mole \space K_2CO_3} = 26.6 \space g \space K_2CO_3$$
b.
1. Find the conversion factors
- 2.50 M $AgNO_3$ solution: $$2.50 \space moles \space AgNO_3 = 1 \space L \space solution$$
- Molar mass: $ AgNO_3 $ : ( 107.9 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 169.9 g/mol
- 1 L = 1000 mL
2. Calculate the mass in grams:
$$10.5 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{2.50 \space moles \space AgNO_3}{1 \space L \space solution} \times \frac{169.9 \space g \space AgNO_3}{1 \space mole \space AgNO_3} = 4.46 \space g \space AgNO_3$$
c.
1. Find the conversion factors
- 6.00 M $H_3PO_4$ solution: $$6.00 \space moles \space H_3PO_4 = 1 \space L \space solution$$
- Molar mass: $ H_3PO_4 $ : ( 1.008 $\times$ 3 )+ ( 16.00 $\times$ 4 )+ ( 30.97 $\times$ 1 )= 97.99 g/mol
- 1 L = 1000 mL
2. Calculate the mass in grams:
$$28.4 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{6.00 \space moles \space H_3PO_4}{1 \space L \space solution} \times \frac{97.99 \space g \space H_3PO_4}{1 \space mole \space H_3PO_4} = 16.7 \space g \space H_3PO_4$$
