Question

given the following chemical reaction:
$2\text{al} + 3\text{h}_2\text{so}_4 \
ightarrow \text{al}_2(\text{so}_4)_3 + 3\text{h}_2$
how many grams of aluminum sulfate, $\text{al}_2(\text{so}_4)_3$, will be produced if 250 g of $\text{h}_2\text{so}_4$ reacts with the aluminum?
move the correct numerical answer to the first box and the correct unit to the second box.
250 270 290 m gr g

Explanation:

Step1: Calculate moles of \( H_2SO_4 \)

Molar mass of \( H_2SO_4 \) is \( 2(1) + 32 + 4(16) = 98 \, \text{g/mol} \). Moles \( = \frac{250 \, \text{g}}{98 \, \text{g/mol}} \approx 2.551 \, \text{mol} \).

Step2: Relate moles of \( H_2SO_4 \) to \( Al_2(SO_4)_3 \)

From reaction, 3 moles \( H_2SO_4 \) produce 1 mole \( Al_2(SO_4)_3 \). Moles of \( Al_2(SO_4)_3 = \frac{2.551}{3} \approx 0.850 \, \text{mol} \).

Step3: Calculate molar mass of \( Al_2(SO_4)_3 \)

Molar mass: \( 2(27) + 3(32 + 4(16)) = 54 + 3(96) = 342 \, \text{g/mol} \).

Step4: Find mass of \( Al_2(SO_4)_3 \)

Mass \( = 0.850 \, \text{mol} \times 342 \, \text{g/mol} \approx 290 \, \text{g} \).

Answer:

290 g