Question

ca(oh)_2 + h_2so_4 = caso_4 + 2h_2o
18.5g 3.2g
how many grams of h_2so_4 were used in the reaction?
13.3
19
88
28

Explanation:

Step1: Determine molar masses

The molar mass of $CaSO_4$ ($M_{CaSO_4}$) is calculated as follows:
$M_{Ca}=40.08\ g/mol$, $M_{S} = 32.07\ g/mol$, $M_{O}=16.00\ g/mol$. So $M_{CaSO_4}=40.08 + 32.07+4\times16.00=136.15\ g/mol$.
The molar mass of $H_2SO_4$ ($M_{H_2SO_4}$) is $M_{H}=1.01\ g/mol$, $M_{S} = 32.07\ g/mol$, $M_{O}=16.00\ g/mol$. So $M_{H_2SO_4}=2\times1.01 + 32.07+4\times16.00 = 98.09\ g/mol$.

Step2: Calculate moles of $CaSO_4$

The number of moles of $CaSO_4$ ($n_{CaSO_4}$) is calculated using the formula $n=\frac{m}{M}$, where $m = 1.85\ g$ and $M = 136.15\ g/mol$. So $n_{CaSO_4}=\frac{1.85\ g}{136.15\ g/mol}\approx0.0136\ mol$.

Step3: Use mole - ratio

From the balanced chemical equation $Ca(OH)_2 + H_2SO_4=CaSO_4 + 2H_2O$, the mole - ratio of $H_2SO_4$ to $CaSO_4$ is $1:1$. So the number of moles of $H_2SO_4$ ($n_{H_2SO_4}$) is equal to the number of moles of $CaSO_4$, i.e., $n_{H_2SO_4}=0.0136\ mol$.

Step4: Calculate mass of $H_2SO_4$

The mass of $H_2SO_4$ ($m_{H_2SO_4}$) is calculated using the formula $m = n\times M$, where $n = 0.0136\ mol$ and $M = 98.09\ g/mol$. So $m_{H_2SO_4}=0.0136\ mol\times98.09\ g/mol\approx1.33\ g$.

Answer:

$1.33$ (assuming the first option in the multiple - choice is $1.33$)