if 1,275 grams of oxygen reacted, how many grams of water were produced? use the following molar masses…

Explanation:

Step1: Calculate moles of oxygen

The formula to calculate the number of moles $n$ is $n=\frac{m}{M}$, where $m$ is the mass and $M$ is the molar - mass. Given $m = 1275$ g of oxygen and $M_{O_2}=32.00$ g/mole. $n_{O_2}=\frac{1275}{32.00}$ moles

Step2: Determine mole - ratio of oxygen to water

The balanced chemical equation for the formation of water from hydrogen and oxygen is $2H_2 + O_2\rightarrow2H_2O$. The mole - ratio of $O_2$ to $H_2O$ is $1:2$. So, the number of moles of water $n_{H_2O}=2\times n_{O_2}=2\times\frac{1275}{32.00}$ moles

Step3: Calculate mass of water

Using the formula $m = n\times M$, with $n = n_{H_2O}$ and $M_{H_2O}=18.02$ g/mole. $m_{H_2O}=2\times\frac{1275}{32.00}\times18.02$ $m_{H_2O}=\frac{2\times1275\times18.02}{32.00}$ $m_{H_2O}=\frac{45972}{32.00}=1436.625$ g

Answer:

$1436.625$ g